Step-by-step explanation:
SiCl₄ + 2 H₂ ----> Si + 4 HCl
According to the problem, 18.1 g of SiCl₄ will react with 8.4 moles of H₂ to give Si and HCl. First we have to determine which of them is the limiting reactant.
To determine the limiting reactant we can find the mass of Si that will be produced by each reactant if they were reacting with an excess of the other one.
According to the coefficients of the reaction, 1 mol of SiCl₄ will produce 1 mol of Si. Let's find the number of moles of Si produced by 18.1 g of SiCl₄
1 mol of SiCl₄ = 1 mol of Si
mass of SiCl₄ = 18.1 g
molar mass of Si = 28.09 g/mol
molar mass of Cl = 35.45 g/mol
molar mass of SiCl₄ = 1 * 28.09 g/mol + 4 * 35.45 g/mol
molar mass of SiCl₄ = 169.89 g/mol
18.1 g of SiCl₄ * (1 mol of SiCl₄/169.89 g) * (1 mol of Si/1 mol of SiCl₄) * (28.09 g of Si/1 mol of Si) = 2.99 g of Si
Now we can do something similar with the other reactant.
moles of H₂ = 8.4 moles
2 moles of H₂ = 1 mol of Si
8.4 moles of H₂ * (1 mol of Si/2 moles of H₂) * (28.09 g of Si/1 mol of Si) = 121.38 g of Si
So SiCl₄ is the limiting reactant and H₂ is in excess.
Answer: 2.99 g of silicone is formed.