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

10.60 grams of silane gas are formed.

Step-by-step 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}} = (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!