Question

A flask in a sealed room that is devoid of oxygen contains 16.05 g of magnesium silicide. An acid HA, where A is a halogen, is added, and a double-replacement reaction occurs. One of the products is a solid, but the other is a gas that escapes from the flask. The mass of the contents of the flask is measured to be 39.85 g. Assuming that the reaction went to completion (that is, neither reactant occurred in excess),
(a) determine the name of the acid.
(b) mathematically prove which product was the gas.

Answer 1

Answer on Question #45803, Chemistry, Inorganic Chemistry

$$M g _ {2} S i + 4 H A \rightarrow 2 M g A _ {2} \downarrow + S i H _ {4} \uparrow$$ $$n (M g _ {2} S i) = \frac {m (M g _ {2} S i)}{M (M g _ {2} S i)} = \frac {1 6 . 0 5}{7 6 . 6 6 6} = 0. 2 0 9 m o l$$

According to reaction equation, $n(MgA_2) = 2n(Mg_2Si) = 0.418mol$. Molar mass of magnesium halogenide is:

$$M (M g A _ {2}) = \frac {m (M g A _ {2})}{n (M g A _ {2})} = \frac {3 9 . 8 5}{0 . 4 1 8} = 9 5. 3 3 g / m o l$$ $$M (A) = \frac {M (M g A _ {2}) - M (M g)}{2} = \frac {9 5 . 3 3 - 2 4 . 3 0 5}{2} = 3 5. 5 g / m o l$$

Therefore, halogen is Chlorine.

According to law of conservation of mass :

$$m (M g _ {2} S i) + m (H C l) = m (M g C l _ {2}) + m (S i H _ {4});$$

According to reaction equation, $n(HCl) = 4n(Mg_2Si) = 0.836mol$.

$$m (H C l) = 0. 8 3 6 m o l * 3 6. 5 g / m o l = 3 0. 5 1 4 g$$ $$m (S i H _ {4}) = m (H C L) + m (M g _ {2} S i) - m (M g C l _ {2}) = 1 6. 0 5 + 3 0. 5 1 4 - 3 9. 8 5 = 6. 7 1 4 g$$ $$n (S i H _ {4}) = \frac {m (S i H _ {4})}{M (S i H _ {4})} = \frac {6 . 7 1 4}{3 2 . 0 8 6} = 0. 2 0 9 m o l$$ $$n (M g _ {2} S i) = n (S i H _ {4})$$

So, gaseous product is silane

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