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 H2Oand SrH2are 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 the question #61228, Chemistry / Inorganic Chemistry

Question:

Strontium Hydride solid reacts with water to form strontium hydroxide and hydrogen gas: $\mathrm{SrH_2(s) + H_2O(L)^*Sr(OH)_2(s) + H_2(g)}$

How many grams of H2O and SrH2 are required to produce 500 grams of $\mathrm{Sr(OH)_2}$? Also, find the volume of hydrogen gas that will be produced from the reaction? Assume the condition is at STP.

Solution:

The reaction equation is:

$$S r H _ {2 (s)} + 2 H _ {2} O _ {(L)} \rightarrow S r (O H) _ {2 (s)} + 2 H _ {2 (g)}$$

The number of the moles ratio for the reaction reagents and products is:

$$n \left(S r H _ {2 (s)}\right) = \frac {n \left(H _ {2} O\right)}{2} = n \left(S r \left(O H\right) _ {2 (s)}\right) = \frac {n \left(H _ {2 (g)}\right)}{2}$$

From the mass, $500\mathrm{g}$ of $Sr(OH)_{2(s)}$, we can calculate the number of the moles of $Sr(OH)_{2(s)}$ and of each substance that took part in the reaction:

$$n \left(S r \left(O H\right) _ {2 (s)}\right) = \frac {m \left(S r \left(O H\right) _ {2 (s)}\right)}{M \left(S r \left(O H\right) _ {2 (s)}\right)} = \frac {5 0 0 (g)}{1 2 1 . 6 3 4 7 \left(\frac {g}{m o l}\right)} = 4. 1 1 m o l.$$

Then, the number of the moles of $SrH_{2(s)}$, used in the reaction, is also equal to 4.11 mol. The number of the moles of $H_{2}O$ used and of $H_{2(g)}$ produced are: $4.11 \cdot 2 = 8.22$ mol.

The mass of $SrH_{2(s)}$ and of $H_{2}O$, required to produce 500g of $Sr(OH)_{2(s)}$, are:

$$\begin{array}{l} m \left(S r H _ {2 (s)}\right) = n \left(S r H _ {2 (s)}\right) \cdot M \left(S r H _ {2 (s)}\right) = 4. 1 1 (m o l) \cdot 8 9. 6 3 5 9 \left(\frac {g}{m o l}\right) = 3 6 9 g \\ m \left(H _ {2} O\right) = n \left(H _ {2} O\right) \cdot M \left(H _ {2} O\right) = 8. 2 2 (m o l) \cdot 1 8. 0 1 5 3 \left(\frac {g}{m o l}\right) = 1 4 8 g \\ \end{array}$$

The volume of 1 mole of gas at STP is 22.4 L. Then, the volume of $H_{2(g)}$ produced is:

$$V \left(H _ {2 (g)}\right) = 8. 2 2 (m o l) \cdot 2 2. 4 \left(\frac {L}{m o l}\right) = 1 8 4 L$$

Answer: The masses of $SrH_{2(s)}$ and of $H_{2}O$, required to produce 500g of $Sr(OH)_{2(s)}$, are 369 g and 148 g, respectively. The volume of hydrogen gas produced, taking STP conditions, is 184 L.

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