How many grams of CaH2CaH2 are needed to generate 147 LL of H2H2 gas if the pressure of H2H2 is 823 torrtorr at 21 ∘C∘C? Express your answer using three significant figures.

139 g of CaH₂ were needed in the reaction

Explanation:

Determine the reaction:

CaH₂ + 2H₂O → Ca (OH) ₂ + 2H₂

1 mol of calcium hidride reacts with 2 moles of water to produce 1 mol of calcium hydroxide and 2 moles of hydrogen

Let's determine the moles of formed hydrogen by the Ideal Gases Law

P. V = n. R. T

P = 823 Torr. 1 atm/760 Torr = 1.08 atm

T = Absolute T° → T°C + 273 → 21°C + 273 = 294K

1.08 atm. 147 L = n. 0.082. 294K

(1.08 atm. 147 L) / (0.082. 294K) = n → 6.60 moles

Ratio is 2:1. We make a rule of three:

2 moles of H₂ came from 1 mol of hydride

Then 6.60 moles of H₂ must came from 3.30 moles of hydride (6.60.1) / 2

Let's convert the moles to mass → 3.30 mol. 42.08 g / 1 mol =

138.8 g ≅ 139 g