Question

What mass in grams of hydrogen is produced by the reaction of 31.3 g of magnesium with 2.12 g of water?

Mg(s) + 2H₂O(l) - Mg(OH)₂ (s) +H₂(g).
a. 0.119
b. 2.71
c. 0.237
d. 0.0593
e. 0.474

Answer 1

The mass of hydrogen produced by the reaction is 1.29 grams.

Step-by-step explanation:

To determine the mass of hydrogen produced, we first need to balance the chemical equation for the reaction:

2Mg(s) + 2H₂O(l) → 2Mg(OH)₂(s) + H₂(g)

From the balanced equation, we can see that for every 2 moles of magnesium (Mg), 1 mole of hydrogen gas (H₂) is produced. Using the molar masses of Mg and H₂, we can calculate the amount of hydrogen gas produced. The molar mass of Mg is 24.305 g/mol, and the molar mass of H₂ is 2.016 g/mol.

Given that we have 31.3 g of Mg, we can set up a conversion:

31.3 g Mg × (1 mol Mg ÷ 24.305 g Mg) × (1 mol H₂ ÷ 2 mol Mg) × (2.016 g H₂ ÷ 1 mol H₂) = 1.29 g H₂

Therefore, the mass of hydrogen produced by the reaction is 1.29 grams (rounded to two decimal places).