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}}](https://tex.z-dn.net/?f=%5Ceta_%7BMg_%7B2%7DSi%7D%20%3D%20%5Cfrac%7Bm_%7BMg_%7B2%7DSi%7D%7D%7BM_%7BMg_%7B2%7DSi%7D%7D)
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](https://tex.z-dn.net/?f=%5Ceta_%7BMg_%7B2%7DSi%7D%20%3D%20%5Cfrac%7Bm_%7BMg_%7B2%7DSi%7D%7D%7BM_%7BMg_%7B2%7DSi%7D%7D%20%3D%20%5Cfrac%7B25.0%20g%7D%7B76.69%20g%2Fmol%7D%20%3D%200.33%20moles)
Now, the stoichiometric relation between Mg₂Si and SiH₄ is 1:1 so:
![\eta_{Mg_{2}Si} = \eta_{SiH_{4}} = 0.33 moles](https://tex.z-dn.net/?f=%20%5Ceta_%7BMg_%7B2%7DSi%7D%20%3D%20%5Ceta_%7BSiH_%7B4%7D%7D%20%3D%200.33%20moles%20)
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](https://tex.z-dn.net/?f=%20m_%7BSiH_%7B4%7D%7D%20%3D%20%5Ceta_%7BSiH_%7B4%7D%7D%2AM_%7BSiH_%7B4%7D%7D%20%3D%200.33%20moles%2A32.12%20g%2Fmol%20%3D%2010.60%20g%20)
Therefore, 10.60 grams of silane gas are formed.
I hope it helps you!