Question

Consider the reaction Mg₂Si(s) + 4 H₂O(ℓ) → 2 Mg(OH)₂(aq) + SiH₄(g). How many grams of silane gas (SiH₄) are formed if 25.0 g of Mg₂Si reacts with excess H₂O?

Answer 1

Answer:

10.60 grams of silane gas are formed.

Explanation:

From the reaction:

Mg₂Si(s) + 4H₂O(l) → 2Mg(OH)₂(aq) + SiH₄(g)          

We know that the limiting reactant is Mg₂Si, so to find the mass of SiH₄ formed we need to calculate the number of moles of Mg₂Si:

$\eta_{Mg_{2}Si} = \frac{m_{Mg_{2}Si}}{M_{Mg_{2}Si}}$

Where:

m: is the mass of Mg₂Si = 25.0 g

M: is the molar mass of Mg₂Si = 76.69 g/mol

$\eta_{Mg_{2}Si} = \frac{m_{Mg_{2}Si}}{M_{Mg_{2}Si}} = \frac{25.0 g}{76.69 g/mol} = 0.33 moles$

Now, the stoichiometric relation between Mg₂Si and SiH₄ is 1:1 so:

$\eta_{Mg_{2}Si} = \eta_{SiH_{4}} = 0.33 moles$

Finally, the mass of SiH₄ is:

$m_{SiH_{4}} = \eta_{SiH_{4}}*M_{SiH_{4}} = 0.33 moles*32.12 g/mol = 10.60 g$

Therefore, 10.60 grams of silane gas are formed.

I hope it helps you!