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

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