Calcium hydride reacts with water to form calcium hydroxide and hydrogen gas. The balanced equation to this is $C a H_{2} + 2 H_{2} O$ $CaH_2+2H_2O$ ---- $C a {(O H)}_{2} + 2 H_{2}$ $Ca(OH)_2+2H_2$ How many grams of calcium hydride are needed to form 8.400 g of hydrogen?

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Answer 1 · 2016-04-01T01:48:32

You need ${87.69 g of CaH}_{2}$ $"87.69 g of CaH"_2$ to form ${8.400 g of H}_{2}$ $"8.400 g of H"_2$ .

There are four steps to answering this type of stoichiometry problem.

Write the balanced equation for the reaction.
Use the molar mass of $H_{2}$ $"H"_2$ to convert grams of $H_{2}$ $"H"_2$ to moles of $H_{2}$ $"H"_2$ .
Use the molar ratio from the balanced equation to convert moles of $H_{2}$ $"H"_2$ to moles of ${CaH}_{2}$ $"CaH"_2$ .
Use the molar mass of ${CaH}_{2}$ $"CaH"_2$ to convert moles of ${CaH}_{2}$ $"CaH"_2$ to grams of ${CaH}_{2}$ $"CaH"_2$ .

Step 1. Write the balanced equation.

${CaH}_{2} + {2H}_{2} O \to {Ca(OH)}_{2} + {2H}_{2}$ $"CaH"_2 + "2H"_2"O" → "Ca(OH)"_2 + "2H"_2$

Step 2. Convert grams of $H_{2}$ $"H"_2$ to moles of $H_{2}$ $"H"_2$ .

The molar mass of $H_{2}$ $"H"_2$ is 2.016 g/mol.

$8.400 cancel("g H"_2) × ("1 mol H"2)/(2.016 cancel("g H"2)) = "4.1667 mol H"_2$

Step 3. Use the molar ratio to calculate the moles of $"CaH"_2$ .

From the balanced equation, the molar ratio is $"1 mol CaH"_2:"2 mol H"_2$ .

$4.1667 cancel("mol H"_2) × ("1 mol CaH"_2)/(2 cancel("mol H"_2)) = "2.0833 mol CaH"_2$

Step 4. Convert moles of $"CaH"_2$ to grams of $"CaH"_2$ .

The molar mass of $"CaH"_2$ is 42.09 g/mol.

$2.0833 cancel("mol CaH"_2) × ("42.09 g CaH"_2)/(1 cancel("mol CaH"_2)) = "87.69 g CaH"2$

The reaction requires $"87.69 g CaH"_2$ .

This video link allows you to furthur practice similar problems.

Calcium hydride reacts with water to form calcium hydroxide and hydrogen gas. The balanced equation to this is CaH_2+2H_2OCaH2+2H2OCaH_2+2H_2O----Ca(OH)_2+2H_2Ca(OH)2+2H2Ca(OH)_2+2H_2 How many grams of calcium hydride are needed to form 8.400 g of hydrogen?