Question

The reaction of calcium hydride with water can be used to prepare small quantities of hydrogen gas, as is done to fill weather-observation balloons. \(\mathrm{CaH}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow\) $$ \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{g}) \text { (not balanced) } $$ (a) How many grams of \(\mathrm{H}_{2}(\mathrm{g})\) result from the reaction of \(127 \mathrm{g} \mathrm{CaH}_{2}\) with an excess of water? (b) How many grams of water are consumed in the reaction of \(56.2 \mathrm{g} \mathrm{CaH}_{2} ?\) (c) What mass of \(\mathrm{CaH}_{2}(\mathrm{s})\) must react with an excess of water to produce \(8.12 \times 10^{24}\) molecules of \(\mathrm{H}_{2} ?\)

Short answer

Answer: (a) 12.16 g of H2 will be produced. (b) 48.11 g of water will be consumed. (c) 283.81 g of CaH2 will be needed.

Step-by-step solution

Balance the chemical equation

The given chemical reaction is unbalanced. To balance it, compare the number of atoms on both sides and adjust the coefficients as necessary. The balanced chemical reaction is: \[ \mathrm{CaH}_{2}(\mathrm{s}) + 2\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s}) + 2\mathrm{H}_{2}(\mathrm{g}) \]

Determine the Molar Mass and carry out Conversions

(a) The molar mass of CaH2 is 42.094 g/mol. So, 127g of CaH2 is \(127 \, \mathrm{g} \times \left( \frac{1 \, \mathrm{mol}}{42.094 \, \mathrm{g}} \right) = 3.01 \, \mathrm{mol}\) Given that the mole ratio of CaH2 to H2 is 1:2, the moles of H2 produced is \(3.01 \, \mathrm{mol} \times 2 = 6.02 \, \mathrm{mol}\). The mass of H2 (with a molar mass of 2.02 g/mol) is then \(6.02 \, \mathrm{mol} \times \left( \frac{2.02 \, \mathrm{g}}{1 \, \mathrm{mol}} \right) = 12.16 \, \mathrm{g}\). (b) The molar mass of water (H2O) is 18.02 g/mol. For 56.2 g of CaH2, the moles is \(56.2 \, \mathrm{g} \times \left( \frac{1 \, \mathrm{mol}}{42.094 \, \mathrm{g}} \right) = 1.33 \, \mathrm{mol}\). The 1:2 mole ratio of CaH2 to H2O then gives \(2 \times 1.33 = 2.67 \, \mathrm{mol} \, H2O\). The mass of water consumed is thus \(2.67 \, \mathrm{mol} \times \left( \frac{18.02 \, \mathrm{g}}{1 \, \mathrm{mol}} \right) = 48.11 \, \mathrm{g}\). (c) The given number of H2 molecules can be converted to moles using Avogadro's number (6.02 x 10^23 molecules/mol); hence \(8.12 \times 10^{24} \, \mathrm{molecules} \times \left( \frac{1 \, \mathrm{mol}}{6.02 \times 10^{23} \, \mathrm{molecules}} \right) = 13.48 \, \mathrm{mol}\). Since the mole ratio of CaH2 to H2 is 1:2, the moles of CaH2 need is \(13.48/2 = 6.74 \, \mathrm{mol}\). Lastly, this is converted to mass using the molar mass of CaH2: \(6.74 \, \mathrm{mol} \times \left( \frac{42.094 \, \mathrm{g}}{1 \, \mathrm{mol}} \right) = 283.81 \, \mathrm{g} \)

Key concepts

Chemical Equations

Chemical equations are like recipes in chemistry. They tell us what reactants are needed and what products are formed. In our exercise, we have the reaction of calcium hydride, CaH extsubscript{2}, with water, H extsubscript{2}O, to produce calcium hydroxide, Ca(OH) extsubscript{2}, and hydrogen gas, H extsubscript{2}. Balancing the equation is crucial because it ensures that the same number of each type of atom is present on both sides. This follows the Law of Conservation of Mass. For the given process, balancing results in: \[ \text{CaH}_2(\text{s}) + 2\, \text{H}_2 \text{O}(\text{l}) \rightarrow \text{Ca}(\text{OH})_2(\text{s}) + 2\, \text{H}_2(\text{g}) \]This tells us that one mole of calcium hydride reacts with two moles of water to produce one mole of calcium hydroxide and two moles of hydrogen gas. Understanding and balancing these equations allow us to determine how much of each substance is needed or produced in a reaction.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It acts like a bridge between the atomic scale and the macroscopic world. For example, the molar mass of calcium hydride (CaH extsubscript{2}) is 42.094 g/mol.

• To find moles from grams, divide the given mass by the molar mass.
• For calcium hydride, 127 g is equal to \( 127\, \text{g} \times \left( \frac{1 \, \text{mol}}{42.094 \, \text{g}} \right) = 3.01 \text{ mol} \).
• Molar mass of hydrogen (H extsubscript{2}) is 2.02 g/mol, and for water (H extsubscript{2}O) it's 18.02 g/mol.

Using these values helps us convert between mass and moles, which is key to solving stoichiometry problems. By understanding molar mass, we can predict how much of a substance will react or be produced.

Gas Production

In chemical reactions, gases can be produced as products and must be quantified accurately. In our exercise, hydrogen gas (H extsubscript{2}) is produced. The equation shows us that for every mole of calcium hydride that reacts, two moles of hydrogen gas are released.

• We use Avogadro’s number \((6.02 \times 10^{23} \text{ molecules/mol})\) to convert between molecules and moles.
• As seen in the example, \( 8.12 \times 10^{24} \) molecules of H extsubscript{2} were converted to moles: \( 8.12 \times 10^{24} \times \left( \frac{1 \, \text{mol}}{6.02 \times 10^{23} \, \text{molecules}} \right) = 13.48 \text{ mol} \).
• The balanced equation and mole ratio help us determine that to produce this amount of hydrogen, \( 6.74 \text{ mol} \) of CaH extsubscript{2} is needed.

Understanding gas production is vital in industries and laboratories, as accurate measurements ensure efficiency and safety in reactions.