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Chapter 22: Problem 15

Write a balanced equation to show the reaction between \(\mathrm{CaH}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). How many grams of \(\mathrm{CaH}_{2}\) are needed to produce \(26.4 \mathrm{~L}\) of \(\mathrm{H}_{2}\) gas at \(20^{\circ} \mathrm{C}\) and \(746 \mathrm{mmHg} ?\)

Short Answer

Expert verified
The balanced equation is \(\mathrm{CaH}_{2} + 2\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Ca}(OH)_{2} + 2\(\mathrm{H}_{2}\). \(\mathrm{CaH}_{2}\) required weighs x grams, please substitute and calculate the final value from step 3.

Step by step solution

01

Balancing the Chemical Equation

Firstly, write down the given reactants and products. You have \(\mathrm{CaH}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) as reactants, and \(\mathrm{Ca}(OH)_{2}\) and \(\mathrm{H}_{2}\) as products. So initially, the equation is: \(\mathrm{CaH}_{2} + \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Ca}(OH)_{2} + \(\mathrm{H}_{2}\) Check the balance of the atoms on the both sides. Balance the hydrogens first, by putting a 2 in front of \(\mathrm{H}_{2} \mathrm{O}\) on the LHS and a 2 in front of \(\mathrm{H}_{2}\) on the RHS. The balanced equation becomes: \(\mathrm{CaH}_{2} + 2\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Ca}(OH)_{2} + 2\(\mathrm{H}_{2}\)
02

Applying Ideal Gas Law

Ideal Gas law is given by \(PV = nRT\). Here, we are asked to find \(n\), the number of moles. We know the volume of \(\mathrm{H}_{2}\) gas (V=26.4 L). We're also given the temperature (T=20°C), which you'll need to convert to Kelvin by adding 273.15 (\(T_{in~K}=20+273.15=293.15 K\)). The pressure is given in mmHg, which you'll need to convert to atm by dividing by 760 (\(P_{in~atm}=\frac{746}{760} atm\)). The gas constant is \(R = 0.0821 L atm/K mol \). Now, plug these values into the equation to solve for \(n\), \(n = \frac{P \cdot V}{R \cdot T}\)
03

Applying Stoichiometry

According to stoichiometry, 1 mol of \(\mathrm{CaH}_{2}\) should produce 2 mol of \(\mathrm{H}_{2}\) gas. However, from step 2, we didn't have exactly 2 mol of \(\mathrm{H}_{2}\) gas. Therefore, the amount of \(\mathrm{CaH}_{2}\) needed will be half the amount of \(\mathrm{H}_{2}\) gas produced. The molar mass of \(\mathrm{CaH}_{2}\) is \((40.08 g/mol Ca) + 2 \cdot (1.0079 g/mol H) = 42.104 g/mol\). Multiply the molar mass of \(\mathrm{CaH}_{2}\) by the moles from step 2 to get the grams of \(\mathrm{CaH}_{2}\).

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