Video: Calculating the Mass of Hydrogen and Oxygen Required to Produce 72 Grams of Water, Given the Balanced Chemical Reaction Equation

What is the mass of hydrogen and oxygen required to produce 72 grams of water according to the equation below? 2H₂ + O₂ ⟶ 2H₂O [A] 1 gram hydrogen gas and 71 grams of oxygen gas [B] 2 grams hydrogen gas and 70 grams of oxygen gas [C] 4 grams hydrogen gas and 68 grams of oxygen gas [D] 6 grams hydrogen gas and 66 grams of oxygen gas [E] 8 grams hydrogen gas and 64 grams of oxygen gas

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Video Transcript

What is the mass of hydrogen and oxygen required to produce 72 grams of water according to the equation below? 2H₂ plus O₂ arrow 2H₂O.

To answer this question best, we need to focus on the stoichiometry of the balanced equation shown in this question. Stoichiometry is the relationship between the relative quantities of substances taking part in a chemical reaction. According to the balanced equation shown in the question, we see that two moles of hydrogen always reacts with one mole of oxygen to produce two moles of water.

Two moles of hydrogen will have a mass in grams equivalent to two times the 𝑀r of the H₂ molecule, where 𝑀r is the relative molecular mass of hydrogen in this particular case. Likewise, one mole of oxygen, or O₂, will have a mass in grams of one times the 𝑀r for the O₂ molecule. And finally, two moles of water will have a mass in grams equivalent to two times the 𝑀r of the H₂O molecule.

So, to calculate these masses using the relative molecular mass of each molecule in the equation, we need to use the relative atomic mass of the atoms concerned. The relative atomic mass of hydrogen is approximately one. And the relative atomic mass of oxygen is approximately 16. This means an oxygen atom is roughly 16 times the mass of a hydrogen atom.

Using these relative atomic mass values correctly in the formulae for hydrogen, oxygen, and water, we can see that four grams of hydrogen molecules always reacts with 32 grams of oxygen molecules to produce 36 grams of water molecules. In the question, we can see that 72 grams of water is actually required. Because 72 grams is precisely double the mass of water from our balanced equation, we must also double the other quantities to keep the reacting mass ratios exactly the same.

By applying this scanning factor to all the reacting masses in the balanced equation, we are maintaining the ratio of the reacting masses. We find that eight grams of hydrogen should react with 64 grams of oxygen to produce 72 grams of water. Eight grams of hydrogen gas and 64 grams of oxygen gas is, therefore, the correct answer.

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