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3 years ago

What is the maxiumum volume of H2 gas at STP that can be produced when 14.0 g of CaH2 and 28.0 g H2O react?

1 answer:

6 0

You are given the mass of CaH2 which is 14 grams and H2O which is 28 grams. You are required to find the maximum volume of H2 gas at STP. The balanced chemical reaction is CaH2 + 2H2O -> Ca(OH)2 + H2. Note that for every one mole of CaH2, 2 moles of H2O is needed to completely react and produce Ca(OH)2 + H2. So the CaH2 and H2O are at equal proportions. At STP, temperature is at 0°C(273K) and pressure at 1 atm. The molar mass of CaH2 is 42 grams per mole.

14g CaH2(1mol CaH2/42 grams CaH2)(1mol H2/1mol CaH2) = 0.333 moles H2

The ideal gas equation is PV=nRT, with R(gas constant) = 0.08206 L-atm/mol-K. Get the equation for volume, we have

V=nRT/P
V=(0.333molH2)( 0.08206 L-atm/mol-K)(273K)/1atm
V=7.47L

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$\boxed{\text{2408 min}}$

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$\begin{array}{rcl}\ln \dfrac{100}{90} & = & k \times 366\\\\1.054 & = & 366k\\\\k & = & \dfrac{1.054 }{366}\\\\k & = & 2.879 \times 10^{-4} \text{ min}^{-1}\\\end{array}\\\\$

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