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 are needed to generate 48.0 L of H₂ gas at a pressure of 0.888 atm and a temperature of 32°C?

Question

Quizplus · Accepted Answer

We can use the Ideal Gas Law to solve for the number of moles of hydrogen gas produced, and then use stoichiometry to determine the required mass of CaH2.

n = PV/RT = (0.888 atm) (48.0 L) / (0.0821 L·atm/mol·K)(305 K) = 1.87 mol H2

From the balanced equation, we see that 1 mole of CaH2 produces 2 moles of H2 gas. Therefore, the number of moles of CaH2 required is:

1.87 mol H2 × (1 mol CaH2/2 mol H2) = 0.935 mol CaH2

Finally, we can convert moles of CaH2 to grams using its molar mass:

0.935 mol CaH2 × 42.1 g/mol = 39.3 g CaH2

Therefore, the answer is D) 35.8 g (rounded to the nearest tenth).

Calcium Hydride (CaH2)reacts with Water to Form Hydrogen Gas

Calcium Hydride (CaH₂)reacts with Water to Form Hydrogen Gas