Question

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?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?35.850.70.85171.7143

Answer 1

Answer: The mass of calcium hydride needed is 35.8 grams

Explanation:

To calculate the moles of hydrogen gas, we use the equation given by ideal gas, which follows:

$PV=nRT$

where,

P = pressure of the hydrogen gas = 0.888 atm

V = Volume of the hydrogen gas = 48.0 L

T = Temperature of the hydrogen gas = $32^oC=[32+273]K=305K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of the hydrogen gas = ?

Putting values in above equation, we get:

$0.888atm\times 48.0L=n_{H_2}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 305K\\\\n_{H_2}=\frac{0.888\times 48.0}{0.0821\times 305}=1.702mol$

The given chemical equation follows:

$CaH_2(s)+2H_2O(l)\rightarrow Ca(OH)_2(aq.)+2H_2$

By Stoichiometry of the reaction:

2 moles of hydrogen gas is produced by 1 mole of calcium hydride

So, 1.702 moles of hydrogen gas will be produced from = $\frac{1}{2}\times 1.702=0.851mol$ of calcium hydride

Now, calculating the mass of calcium hydride by using the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of calcium hydride = 0.851 moles

Molar mass of calcium hydride = 42 g/mol

Putting values in above equation, we get:

$0.851mol=\frac{\text{Mass of calcium hydride}}{42g/mol}\\\\\text{Mass of calcium hydride}=(0.851mol\times 42g/mol)=35.8g$

Hence, the mass of calcium hydride needed is 35.8 grams