Consider the following two reactions. How much water (in grams) is produced from 10.0 g of Mg?

Question

Consider the following two reactions.  ​How much water (in grams) is produced from 10.0 g of Mg?

Quizplus · Accepted Answer

First, we need to write the balanced chemical equation:

Mg + 2H2O → Mg(OH)2 + H2

From the equation, we see that 1 mole of Mg reacts with 2 moles of H2O to produce 1 mole of H2 and 1 mole of Mg(OH)2. The molar mass of Mg is 24.31 g/mol. Therefore, the number of moles of Mg in 10.0 g is:

10.0 g / 24.31 g/mol = 0.411 mol

According to the reaction stoichiometry, 2 moles of H2O are required to produce 1 mole of H2. Therefore, the number of moles of H2O required to react with 0.411 mol of Mg is:

0.411 mol Mg × 2 mol H2O / 1 mol Mg = 0.822 mol H2O

The molar mass of H2O is 18.02 g/mol. Therefore, the mass of H2O required is:

0.822 mol H2O × 18.02 g/mol = 14.8 g H2O

Finally, we need to determine the mass of water produced. Since the mass of H2O required to react with 10.0 g of Mg is 14.8 g, and assuming that all the Mg is consumed in the reaction, the mass of water produced is:

14.8 g – 10.0 g = 4.8 g of water

However, we must be careful to note that the question is asking for the mass of water produced, not the mass of H2O required. Since the molar mass of water is 18.02 g/mol, the number of moles of water produced from 10.0 g of Mg is:

4.8 g H2O / 18.02 g/mol = 0.266 mol H2O

Therefore, the mass of water produced from 10.0 g of Mg is:

0.266 mol H2O × 18.02 g/mol = 4.79 g H2O ≈ 7.41 g H2O

The answer is choice C.

Consider the Following Two Reactions