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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 0.995 atm and a temperature of 32 °C?

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Answer:

40.g CaH2

Explanation:

1. ideal gas law(PV = nRT) → use ideal gas law first when volume is given

P = 0.995atm

V = 48.0L H2

n = ?

R = 0.0821L atm/molK

T = 32 + 273 = 305K

n = (0.995atm x 48.0L H2)/(0.0821L atm/molK x 305K) → do not simplify as small decimals might change the answer

2. Conversions

2 H2 and 1 CaH2 → 1/2

(mole of H2) x 1/2 x (molar mass of CaH2)

(0.995atm x 48.0L H2)/(0.0821L atm/molK x 305K) x 1/2 x (40.08 + 2.02) = 40.g CaH2

the longer answer will be 40.14887882 but as the minimum sigfig given in the question is 2, it is 40.g CaH2.

Hope it helped!

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