The industrial synthesis of ammonia, which feeds half the world's agriculture, can be summed up in one line: $\text{N}_2 + 3\ \text{H}_2 \rightarrow 2\ \text{NH}_3$. The numbers in front of the formulae are not put there at random: they count, atom by atom, everything that goes into the reactor and everything that comes out. You balanced simple equations last year; now it is time to do it with longer equations, without being afraid of fractions.
Drinking water ($\text{H}_2\text{O}$) is essential for life, but hydrogen peroxide ($\text{H}_2\text{O}_2$) burns the skin and is toxic if swallowed. Why does changing a small number create a different chemical substance, while changing the big number in front only tells you how much you have in the glass?
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Count the atoms of each element on both sides of equation $\text{CH}4 + \text{O}2 \rightarrow \text{CO}2 + \text{H}2\text{O}$, as it is written. The figure shows it…
Count the atoms of each element on both sides of equation $\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$, as it is written. The figure shows it using the particle model:
Reason whether it is balanced or not. If it is not, state which elements fail.
Hydrochloric acid is obtained industrially by reacting gaseous hydrogen and chlorine. Balance the equation: $\text{H}2 + \text{Cl}2 \rightarrow \text{HCl}$.
Hydrochloric acid is obtained industrially by reacting gaseous hydrogen and chlorine. Balance the equation: $\text{H}_2 + \text{Cl}_2 \rightarrow \text{HCl}$.
If you try to balance the combustion reaction of a match without any fixed order, you can spend ten minutes changing numbers from one side to the other and watching the oxygen become unbalanced again and again. Is there a systematic strategy for balancing any equation first time without getting stuck in endless loops?
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Metallic calcium reacts with oxygen in the air to form calcium oxide, quicklime. Balance the equation: $\text{Ca} + \text{O}2 \rightarrow \text{CaO}$.
Metallic calcium reacts with oxygen in the air to form calcium oxide, quicklime. Balance the equation: $\text{Ca} + \text{O}_2 \rightarrow \text{CaO}$.
Sodium is a soft metal that reacts violently with chlorine to form common salt. Balance the equation: $\text{Na} + \text{Cl}2 \rightarrow \text{NaCl}$.
Sodium is a soft metal that reacts violently with chlorine to form common salt. Balance the equation: $\text{Na} + \text{Cl}_2 \rightarrow \text{NaCl}$.
Car engines produce nitrogen monoxide, $\text{NO}$, which oxidises when released into the air to form the brownish nitrogen dioxide seen in pollution episodes. Balance…
Car engines produce nitrogen monoxide, $\text{NO}$, which oxidises when released into the air to form the brownish nitrogen dioxide seen in pollution episodes. Balance the equation: $\text{NO} + \text{O}_2 \rightarrow \text{NO}_2$.
The more substances are involved, the easier it is to unbalance something you had already balanced. That is why it is best to go element by element, in the order from the previous section, and check at the end.
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When baking a cake, the sodium hydrogen carbonate in the baking powder decomposes with heat into sodium carbonate, water and the carbon dioxide that makes the dough…
When baking a cake, the sodium hydrogen carbonate in the baking powder decomposes with heat into sodium carbonate, water and the carbon dioxide that makes the dough rise. Balance the equation: $\text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2$.
The phosphorus on the match head burns with oxygen in the air to form phosphorus(V) oxide. Balance the equation: $\text{P} + \text{O}2 \rightarrow \text{P}2\text{O}5$.
The phosphorus on the match head burns with oxygen in the air to form phosphorus(V) oxide. Balance the equation: $\text{P} + \text{O}_2 \rightarrow \text{P}_2\text{O}_5$.
Lavoisier and Priestley studied oxygen by heating mercury oxide, a red powder that decomposes into liquid mercury and oxygen gas. The figure shows how the particles of…
Lavoisier and Priestley studied oxygen by heating mercury oxide, a red powder that decomposes into liquid mercury and oxygen gas. The figure shows how the particles of each substance are formed:
a) Balance the equation: $\text{HgO} \rightarrow \text{Hg} + \text{O}_2$.
b) Mercury is written as $\text{Hg}$ and not $\text{Hg}_2$, whereas oxygen does have a subscript. Explain, using the particle model, the reason for this difference.
Sometimes, when you balance an element, the number you get is a fraction. This is not a mistake: you just need one more step.
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Ethane, $\text{C}2\text{H}6$, accompanies methane in natural gas. a) Balance its combustion, $\text{C}2\text{H}6 + \text{O}2 \rightarrow \text{CO}2 + \text{H}2\text{O}$,…
Ethane, $\text{C}_2\text{H}_6$, accompanies methane in natural gas.
a) Balance its combustion, $\text{C}_2\text{H}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$, keeping the coefficient of ethane equal to $1$ for now: you will find that the coefficient for oxygen turns out to be a fraction.
b) Then multiply all the coefficients by the appropriate number to write the equation with the lowest possible integer coefficients.
Oxy-acetylene torches in workshops burn acetylene, $\text{C}2\text{H}2$, with oxygen and reach about $3\,000\ \celsius$. a) Balance the combustion equation:…
Oxy-acetylene torches in workshops burn acetylene, $\text{C}_2\text{H}_2$, with oxygen and reach about $3\,000\ \celsius$.
a) Balance the combustion equation: $\text{C}_2\text{H}_2 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$.
b) If a fraction appeared while balancing it, explain how to convert the balanced equation into lowest integer coefficients and write the final equation.
When balancing the combustion of propane, a student wrote: $2\ \text{C}3\text{H}8 + 10\ \text{O}2 \rightarrow 6\ \text{CO}2 + 8\ \text{H}2\text{O}$. a) Check, by…
When balancing the combustion of propane, a student wrote: $2\ \text{C}_3\text{H}_8 + 10\ \text{O}_2 \rightarrow 6\ \text{CO}_2 + 8\ \text{H}_2\text{O}$.
a) Check, by counting atoms, that their equation respects the conservation of atoms.
b) Even so, the teacher marks it as "could be improved". Explain what it means that the coefficients must be the lowest possible integers and rewrite the equation correctly.
In the synthesis of ammonia, $28\ \text{g}$ of nitrogen react with just $6\ \text{g}$ of hydrogen to give exactly $34\ \text{g}$ of product with no gas left over. Why do the substances react in such different masses if the balanced formula involves simple numbers such as $1$, $3$ and $2$?
This reading in moles is the gateway to the calculations with masses that you will see in the next section: the coefficients will remain the key.
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The industrial synthesis of ammonia is written as $\text{N}2 + 3\ \text{H}2 \rightarrow 2\ \text{NH}3$, as shown in the figure using the particle model: Interpret it in…
The industrial synthesis of ammonia is written as $\text{N}_2 + 3\ \text{H}_2 \rightarrow 2\ \text{NH}_3$, as shown in the figure using the particle model:
Interpret it in two ways:
a) Using molecules: how many molecules of nitrogen and hydrogen react and how many molecules of ammonia are formed.
b) Using moles: how many moles of each gas are consumed when $2\ \text{mol}$ of ammonia are formed.
The thin layer that protects the aluminium parts of an aeroplane from the air is aluminium oxide. The figure shows the particles involved: Check, by counting the atoms…
The thin layer that protects the aluminium parts of an aeroplane from the air is aluminium oxide. The figure shows the particles involved:
Check, by counting the atoms of each element on both sides, whether this equation is balanced: $4\ \text{Al} + 3\ \text{O}_2 \rightarrow 2\ \text{Al}_2\text{O}_3$.
In the electrolysis of water, electric current decomposes it into hydrogen and oxygen gases. Balance the equation: $\text{H}2\text{O} \rightarrow \text{H}2 + \text{O}2$.
In the electrolysis of water, electric current decomposes it into hydrogen and oxygen gases. Balance the equation: $\text{H}_2\text{O} \rightarrow \text{H}_2 + \text{O}_2$.
Raising the temperature usually increases the rate because a larger fraction of collisions overcomes the barrier. Increasing the concentration of reactants in solution or in gases usually increases the number of encounters. Crushing a solid increases the contact surface area. A catalyst provides a pathway with lower activation energy: it takes part in intermediate steps and is regenerated; it does not provide net energy or, on its own, increase the equilibrium yield.
Design a comparison using one whole effervescent tablet and one crushed tablet: same mass, water, temperature and container. Repeat and measure the time until the same end point; change only the surface area. Do it under supervision, wearing goggles, without swallowing any products or sealing the container. Record hypotheses, variables, spread and limitations: also changing the temperature would make it impossible to attribute the change to a single factor.
The diagram links the main ideas:
Let's practice! 8 medium exercises in this section