2 CH3Cl + Si --> (CH3)2SiCl2
If 170 g of CH3Cl and 143 g of Si are mixed together, what is the maximum amount (in grams) of (CH3)2SiCl2 that can be formed?
This is a limiting reagent problem. I know that because amounts for BOTH reactants are given. Here is a site that shows, in detail, how to solve limiting reagent problems.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html
To determine the maximum amount of (CH3)2SiCl2 that can be formed, we need to use the concept of limiting reactants. In a chemical reaction, the limiting reactant is the one that is completely consumed first and determines the maximum amount of product that can be produced.
To find the limiting reactant, we need to compare the number of moles of each reactant. We can then use stoichiometry to calculate the moles of (CH3)2SiCl2 that can be formed from each reactant.
First, we need to calculate the number of moles of each reactant using their molar masses:
Molar mass of CH3Cl (methyl chloride) = 12.01 g/mol (C) + 1.01 g/mol (H) + 35.45 g/mol (Cl) = 50.47 g/mol
Molar mass of Si (silicon) = 28.09 g/mol
Moles of CH3Cl = 170 g / 50.47 g/mol = 3.37 mol
Moles of Si = 143 g / 28.09 g/mol = 5.09 mol
Now, we can use the balanced equation to determine the stoichiometric ratio between CH3Cl and (CH3)2SiCl2:
2 CH3Cl + Si → (CH3)2SiCl2
From the balanced equation, we can see that 2 moles of CH3Cl react with 1 mole of Si to produce 1 mole of (CH3)2SiCl2.
Using this ratio, we can calculate the moles of (CH3)2SiCl2 that can be formed from each reactant:
Moles of (CH3)2SiCl2 from CH3Cl = 3.37 mol CH3Cl × (1 mol (CH3)2SiCl2 / 2 mol CH3Cl) = 1.69 mol (CH3)2SiCl2
Moles of (CH3)2SiCl2 from Si = 5.09 mol Si × (1 mol (CH3)2SiCl2 / 1 mol Si) = 5.09 mol (CH3)2SiCl2
Now we compare the moles of (CH3)2SiCl2 formed from each reactant. The smallest value represents the maximum amount of (CH3)2SiCl2 that can be formed:
The maximum amount of (CH3)2SiCl2 that can be formed is 1.69 grams.