Stoichiometric calculations
An engineer designing a reactor at a Gulf Coast refinery does not work with individual molecules; they work with tons. A balanced equation, on its own, only gives a ratio of particles: $1$ molecule of methane for every $2$ of oxygen. Turning that abstract ratio into the kilograms that have to be bought or that will come out of the smokestack requires an essential intermediate step: the mole.
Molar mass: the bridge between the balance and the particle counter
If you put $12\text{ g}$ of carbon on one balance and $12\text{ g}$ of magnesium on another, both show the same mass, but since a magnesium atom has twice the mass of a carbon atom, the second pan contains exactly half as many atoms: how can we count the particles of a substance simply by weighing it?
The balanced equation for the combustion of methane, $\text{CH}4 + 2\ \text{O}2 \rightarrow \text{CO}2 + 2\ \text{H}2\text{O}$, can also be read in moles. a) How many…
The balanced equation for the combustion of methane, $\text{CH}_4 + 2\ \text{O}_2 \rightarrow \text{CO}_2 + 2\ \text{H}_2\text{O}$, can also be read in moles.
- a) How many moles of oxygen are needed to burn $3\ \text{mol}$ of methane?
- b) How many moles of carbon dioxide and how many of water are formed then?
The glucose used by your muscles when playing sports has the formula $\text{C}6\text{H}{12}\text{O}6$. Calculate its molar mass and explain in a sentence what this value…
The glucose used by your muscles when playing sports has the formula $\text{C}_6\text{H}_{12}\text{O}_6$. Calculate its molar mass and explain in a sentence what this value means (atomic masses: $\text{H} = 1\ \text{u}$; $\text{C} = 12\ \text{u}$; $\text{O} = 16\ \text{u}$).
A glass contains $36\ \text{g}$ of water. Knowing that the molar mass of water is $18\ \text{g/mol}$, calculate how many moles of water are in the glass.
A glass contains $36\ \text{g}$ of water. Knowing that the molar mass of water is $18\ \text{g/mol}$, calculate how many moles of water are in the glass.
A flask contains $0.5\ \text{mol}$ of carbon dioxide. Knowing that the molar mass of $\text{CO}2$ is $44\ \text{g/mol}$, calculate the mass of gas contained in the flask.
A flask contains $0.5\ \text{mol}$ of carbon dioxide. Knowing that the molar mass of $\text{CO}_2$ is $44\ \text{g/mol}$, calculate the mass of gas contained in the flask.
The general method: mass → moles → moles → mass
The equation for the formation of water is $2\ \text{H}2 + \text{O}2 \rightarrow 2\ \text{H}2\text{O}$. If $4\ \text{mol}$ of hydrogen is completely burned: a) How many…
The equation for the formation of water is $2\ \text{H}_2 + \text{O}_2 \rightarrow 2\ \text{H}_2\text{O}$. If $4\ \text{mol}$ of hydrogen is completely burned:
- a) How many moles of oxygen are consumed?
- b) How many moles of water are formed?
In the synthesis of ammonia, $\text{N}2 + 3\ \text{H}2 \rightarrow 2\ \text{NH}3$, $6\ \text{mol}$ of hydrogen reacts with sufficient nitrogen. a) How many moles of…
In the synthesis of ammonia, $\text{N}_2 + 3\ \text{H}_2 \rightarrow 2\ \text{NH}_3$, $6\ \text{mol}$ of hydrogen reacts with sufficient nitrogen.
- a) How many moles of nitrogen are consumed?
- b) How many moles of ammonia are obtained?
At a barbecue, $12\ \text{g}$ of carbon burns according to the equation $\text{C} + \text{O}2 \rightarrow \text{CO}2$. Calculate, by converting through moles, what mass…
At a barbecue, $12\ \text{g}$ of carbon burns according to the equation $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$. Calculate, by converting through moles, what mass of carbon dioxide is formed (atomic masses: $\text{C} = 12\ \text{u}$; $\text{O} = 16\ \text{u}$).
A home furnace burns $16\ \text{g}$ of methane according to the equation $\text{CH}4 + 2\ \text{O}2 \rightarrow \text{CO}2 + 2\ \text{H}2\text{O}$. Calculate, by…
A home furnace burns $16\ \text{g}$ of methane according to the equation $\text{CH}_4 + 2\ \text{O}_2 \rightarrow \text{CO}_2 + 2\ \text{H}_2\text{O}$. Calculate, by converting through moles:
- a) The mass of oxygen it consumes.
- b) The mass of carbon dioxide it emits.
(molar masses: $\text{CH}_4 = 16\ \text{g/mol}$; $\text{O}_2 = 32\ \text{g/mol}$; $\text{CO}_2 = 44\ \text{g/mol}$).
You want to obtain $90\ \text{g}$ of water by reacting hydrogen with oxygen according to the equation $2\ \text{H}2 + \text{O}2 \rightarrow 2\ \text{H}2\text{O}$.…
You want to obtain $90\ \text{g}$ of water by reacting hydrogen with oxygen according to the equation $2\ \text{H}_2 + \text{O}_2 \rightarrow 2\ \text{H}_2\text{O}$. Calculate, by converting through moles, what mass of oxygen is needed (molar masses: $\text{O}_2 = 32\ \text{g/mol}$; $\text{H}_2\text{O} = 18\ \text{g/mol}$).
A chemical plant reacts $28\ \text{g}$ of nitrogen with excess hydrogen according to the equation $\text{N}2 + 3\ \text{H}2 \rightarrow 2\ \text{NH}3$. Calculate, by…
A chemical plant reacts $28\ \text{g}$ of nitrogen with excess hydrogen according to the equation $\text{N}_2 + 3\ \text{H}_2 \rightarrow 2\ \text{NH}_3$. Calculate, by converting through moles, the mass of ammonia obtained (molar masses: $\text{N}_2 = 28\ \text{g/mol}$; $\text{NH}_3 = 17\ \text{g/mol}$).
Industrial quantities and impure raw materials
The method does not change with enormous quantities: you just have to watch the units. But real raw material is almost never pure: a piece of limestone may contain impurities that do not react, and only the pure fraction counts in the stoichiometric calculation.
In a lime kiln, $100\ \text{g}$ of calcium carbonate decomposes according to the equation $\text{CaCO}3 \rightarrow \text{CaO} + \text{CO}2$. Calculate, by converting…
In a lime kiln, $100\ \text{g}$ of calcium carbonate decomposes according to the equation $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$. Calculate, by converting through moles, the mass of quicklime ($\text{CaO}$) obtained (molar masses: $\text{CaCO}_3 = 100\ \text{g/mol}$; $\text{CaO} = 56\ \text{g/mol}$).
A quarry supplies a lime factory with limestone containing $80\ \%$ calcium carbonate (the remainder is unreactive impurities). $250\ \text{g}$ of this rock is loaded…
A quarry supplies a lime factory with limestone containing $80\ \%$ calcium carbonate (the remainder is unreactive impurities). $250\ \text{g}$ of this rock is loaded into the kiln: $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$. The bar shows the composition of the rock:
- a) What mass of calcium carbonate actually enters the kiln?
- b) Calculate, by converting through moles, the mass of quicklime ($\text{CaO}$) obtained.
(molar masses: $\text{CaCO}_3 = 100\ \text{g/mol}$; $\text{CaO} = 56\ \text{g/mol}$).
A steel mill receives iron ore with $80\ \%$ iron(III) oxide (the remainder is inert gangue). In the blast furnace, the oxide reacts according to…
A steel mill receives iron ore with $80\ \%$ iron(III) oxide (the remainder is inert gangue). In the blast furnace, the oxide reacts according to $\text{Fe}_2\text{O}_3 + 3\ \text{CO} \rightarrow 2\ \text{Fe} + 3\ \text{CO}_2$. $200\ \text{g}$ of ore is loaded:
- a) What mass of $\text{Fe}_2\text{O}_3$ does that charge contain?
- b) Calculate, by converting through moles, the mass of iron obtained.
(molar masses: $\text{Fe}_2\text{O}_3 = 160\ \text{g/mol}$; $\text{Fe} = 56\ \text{g/mol}$).
The limiting reactant: the mole quotient decides
In a container, $10\ \text{g}$ of hydrogen is mixed with $64\ \text{g}$ of oxygen and a spark is ignited: $2\ \text{H}2 + \text{O}2 \rightarrow 2\ \text{H}2\text{O}$.…
In a container, $10\ \text{g}$ of hydrogen is mixed with $64\ \text{g}$ of oxygen and a spark is ignited: $2\ \text{H}_2 + \text{O}_2 \rightarrow 2\ \text{H}_2\text{O}$. The diagram summarizes the experiment data:
- a) Calculate the moles of each reactant and deduce which one is in excess.
- b) What mass of water is formed?
- c) What mass of excess reactant remains unreacted?
(molar masses: $\text{H}_2 = 2\ \text{g/mol}$; $\text{O}_2 = 32\ \text{g/mol}$; $\text{H}_2\text{O} = 18\ \text{g/mol}$).
$14\ \text{g}$ of nitrogen and $6\ \text{g}$ of hydrogen are introduced into a reactor to produce ammonia: $\text{N}2 + 3\ \text{H}2 \rightarrow 2\ \text{NH}3$. a)…
$14\ \text{g}$ of nitrogen and $6\ \text{g}$ of hydrogen are introduced into a reactor to produce ammonia: $\text{N}_2 + 3\ \text{H}_2 \rightarrow 2\ \text{NH}_3$.
- a) Calculate the moles of each reactant and deduce which one is used up first.
- b) What mass of ammonia is obtained?
- c) What mass of the other reactant is left over?
(molar masses: $\text{N}_2 = 28\ \text{g/mol}$; $\text{H}_2 = 2\ \text{g/mol}$; $\text{NH}_3 = 17\ \text{g/mol}$).
In a furnace, $32\ \text{g}$ of methane is burned with $96\ \text{g}$ of oxygen: $\text{CH}4 + 2\ \text{O}2 \rightarrow \text{CO}2 + 2\ \text{H}2\text{O}$. a) Calculate…
In a furnace, $32\ \text{g}$ of methane is burned with $96\ \text{g}$ of oxygen: $\text{CH}_4 + 2\ \text{O}_2 \rightarrow \text{CO}_2 + 2\ \text{H}_2\text{O}$.
- a) Calculate the moles of each reactant and deduce which one is the limiting reactant.
- b) What mass of carbon dioxide is emitted?
- c) What mass of oxygen is left over?
(molar masses: $\text{CH}_4 = 16\ \text{g/mol}$; $\text{O}_2 = 32\ \text{g/mol}$; $\text{CO}_2 = 44\ \text{g/mol}$).
Theoretical, actual and percent yield
The mass of product you calculate is the theoretical yield: the most you could get if everything reacted perfectly. In a real reactor, some product is lost or side reactions happen, so the mass you actually collect, the actual yield, is smaller.
When calcining $100\ \text{g}$ of pure calcium carbonate, $\text{CaCO}3 \rightarrow \text{CaO} + \text{CO}2$, theory states that $56\ \text{g}$ of quicklime should be…
When calcining $100\ \text{g}$ of pure calcium carbonate, $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$, theory states that $56\ \text{g}$ of quicklime should be obtained. However, a real kiln has only produced $42\ \text{g}$, as compared by the bars in the figure.
- a) Calculate the percentage yield of the kiln.
- b) Propose a reason explaining why the kiln does not reach $100\ \%$.
According to equation $\text{N}2 + 3\ \text{H}2 \rightarrow 2\ \text{NH}3$, $28\ \text{g}$ of nitrogen should yield $34\ \text{g}$ of ammonia. In a real reactor, only…
According to equation $\text{N}_2 + 3\ \text{H}_2 \rightarrow 2\ \text{NH}_3$, $28\ \text{g}$ of nitrogen should yield $34\ \text{g}$ of ammonia. In a real reactor, only $20.4\ \text{g}$ has been collected.
- a) Calculate the percent yield of the reaction.
- b) Explain two reasons why a factory might not achieve the theoretical yield of $100\ \%$.
Back to the beginning: To burn $1\text{ kg}$ of methane ($16\text{ g/mol}$, $62.5\text{ mol}$), the $1:2$ ratio requires $125\text{ mol}$ of $\text{O}_2$ ($32\text{ g/mol}$): the engineer must supply exactly $4\text{ kg}$ of oxygen for every kilogram of gas.