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Wittaler [7]
3 years ago
14

Calcium hydride (cah2) reacts with water to form hydrogen gas: cah2(s) + 2h2o(l) → ca(oh)2(aq) + 2h2(g) how many grams of cah2 a

re needed to generate 58.0 l of h2 gas at a pressure of 0.811 atm and a temperature of 32°c?
Chemistry
1 answer:
andreev551 [17]3 years ago
3 0
Using the ideal gas law equation, we can find the number of H₂ moles produced.
PV = nRT
Where P - pressure - 0.811 atm x 101 325 Pa/atm = 82 175 Pa
V - volume - 58.0 x 10⁻³ m³
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 32 °C + 273 = 305 K
substituting these values in the equation,
82 175 Pa x 58.0 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 305 K
n = 1.88 mol 

The balanced equation for the reaction is as follows;
CaH₂(s) + 2H₂O(l) --> Ca(OH)₂(aq) + 2H₂(g)
stoichiometry of CaH₂ to H₂ is 1:2
When 1.88 mol of H₂ is formed , number of CaH₂ moles reacted = 1.88/2 mol
therefore number of CaH₂ moles reacted = 0.94 mol
Mass of CaH₂ reacted - 0.94 mol x 42 g/mol  = 39.48 g of CaH₂ are needed
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<h3>Answer:</h3>

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<h3>Explanation:</h3>

<u>We are given;</u>

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But, 1000 mL = 1 L

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We are required to calculate the molarity of the solution;

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In this case;

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