Question

2. Strontium Hydride solid reacts with water to form strontium hydroxide and hydrogen gas:
SrH2(s) + H2O(L) -> Sr(OH)2(s) + H2(g)
How many grams of H2O and SrH2 are required to produce 500 grams of Sr(OH)2? Also, find the volume of hydrogen gas that will be produced from the reaction? Assume the condition is at STP.

Answer 1

Answer on question #61160, Chemistry / General Chemistry

Strontium Hydride solid reacts with water to form strontium hydroxide and hydrogen gas: $\mathrm{SrH2(s) + H2O(L) \rightarrow Sr(OH)2(s) + H2(g)}$.

How many grams of H2O and SrH2 are required to produce 500 grams of Sr(OH)2? Also, find the volume of hydrogen gas that will be produced from the reaction? Assume the condition is at STP.

**Solution:**

\begin{array}{l} \mathrm{SrH_2(s) + 2H_2O(L) \rightarrow Sr(OH)_2(s) + 2H_2(g)} \\ \mathrm{M(SrH_2) = 89.6\ g/mol;} \\ \mathrm{M(H_2O) = 18 \times 2 = 36\ g/mol\ (because 2\ mol\ H_2O).};} \\ \mathrm{M(Sr(OH)_2) = 121.6\ g/mol;} \\ \mathrm{M(H_2) = 2 \times 2 = 4\ g/mol\ (because 2\ mol\ H_2).} \end{array}

We draw up proportion:

$$x\ (g) : 89.6\ (g/mol) = 500\ (g) : 121.6\ (g/mol)$$

Where x - the mass of $\mathrm{SrH_2}$

$$x = \frac{89.6\ \left(\frac{g}{mol}\right) \times 500\ (g)}{121.6\ \left(\frac{g}{mol}\right)} = 368.42\ g$$ $$m(\mathrm{SrH_2}) = 368.42\ (g).$$ $$x\ (g) : 36\ (g/mol) = 500\ (g) : 121.6\ (g/mol)$$

Where x - the mass of $\mathrm{H_2O}$

$$x = \frac{36\ \left(\frac{g}{mol}\right) \times 500\ (g)}{121.6\ \left(\frac{g}{mol}\right)} = 148.02\ g$$ $$m(\mathrm{H_2O}) = 148.02\ (g).$$

We find how many grams of $\mathrm{H_2}$ can be produced in the reaction.

Using the limiting reactant's value of $\mathrm{SrH_2}$, because of $\mathrm{H_2O}$ can produce a larger quantity of $\mathrm{H_2}$ than of $\mathrm{SrH_2}$

$$x = \frac{4\ \left(\frac{g}{mol}\right) \times 368.33\ (g)}{89.6\ \left(\frac{g}{mol}\right)} = 16.44\ g$$

Find the volume of $\mathrm{H_2}$.

Mass of 1 mole of hydrogen is 2 g. One mole of any gas will occupy a volume of 22.4 liters at STP.

We draw up proportion:

$$2 (g): 22.4 (L) = 16.44 (g): x (L),$$

Where x - the volume of H₂

$$x = \frac{22.4 (L) \times 16.44 (g)}{2 (g)} = 184.128 L$$ $$V(H_2) = 184.128 (L).$$

Answer: $m(SrH_2) = 368.42 (g)$; $m(H_2O) = 148.02 (g)$; $V(H_2) = 184.128 (L)$