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:
[tex]\eta_{Mg_{2}Si} = \frac{m_{Mg_{2}Si}}{M_{Mg_{2}Si}}[/tex]
Where:
m: is the mass of Mg₂Si = 25.0 g
M: is the molar mass of Mg₂Si = 76.69 g/mol
[tex]\eta_{Mg_{2}Si} = \frac{m_{Mg_{2}Si}}{M_{Mg_{2}Si}} = \frac{25.0 g}{76.69 g/mol} = 0.33 moles[/tex]
Now, the stoichiometric relation between Mg₂Si and SiH₄ is 1:1 so:
[tex] \eta_{Mg_{2}Si} = \eta_{SiH_{4}} = 0.33 moles [/tex]
Finally, the mass of SiH₄ is:
[tex] m_{SiH_{4}} = \eta_{SiH_{4}}*M_{SiH_{4}} = 0.33 moles*32.12 g/mol = 10.60 g [/tex]
Therefore, 10.60 grams of silane gas are formed.
I hope it helps you!
