When 20.5 g of methane and 45.0 g of chlorine gas undergo a reaction that has a 75.0% yield, what mass (g) of chloromethane (CH3Cl) forms? Hydrogen chloride also forms.
CH4 + Cl2 ==> CH3Cl + HCl
mols CH4 = g/molar mass = 20.5/16 = about 1.3 (you need to clean up my estimates).
mols Cl2 = 45/71 = about 0.64g
mols CH3Cl formed from CH4 = 1.28
mols CH3Cl formed from Cl2 = 0.64
Obviously both numbers can't be right; the correct value in limiting reagent problems is ALWAYS the smaller one and the reagent producing that value is the limiting reagent. Thus Cl2 is the limiting reagent and 0.64 mols CH3Cl will be formed.
If you have a 75% yield, the amount ACTUALLY formed will be 0.64 x 0.75 = ?g
To determine the mass of chloromethane (CH3Cl) formed in the given reaction, we need to use the concept of stoichiometry. In this case, we'll use the balanced chemical equation for the reaction of methane (CH4) and chlorine (Cl2) gas to form chloromethane (CH3Cl) and hydrogen chloride (HCl):
CH4 + Cl2 → CH3Cl + HCl
The molar mass of methane is approximately 16.04 g/mol, and the molar mass of chlorine gas is approximately 70.90 g/mol. To find the moles of methane and chlorine gas, we can use the following equations:
moles of methane = mass of methane / molar mass of methane
moles of chlorine gas = mass of chlorine gas / molar mass of chlorine gas
Substituting the given masses of methane and chlorine gas:
moles of methane = 20.5 g / 16.04 g/mol
moles of chlorine gas = 45.0 g / 70.90 g/mol
Next, we can use the stoichiometry of the balanced equation to determine the moles of chloromethane produced. From the balanced equation:
1 mol of methane reacts with 1 mol of chlorine gas to form 1 mol of chloromethane
Therefore, the moles of chloromethane formed will be equal to the lesser of the moles of methane and chlorine gas. We'll assume that methane is the limiting reactant, so the moles of chloromethane formed will be equal to:
moles of chloromethane = moles of methane
Next, we need to consider the yield of the reaction. The yield is given as 75%, which means that only 75% of the expected product is obtained. To find the actual moles of chloromethane formed, we can multiply the theoretical moles (from the stoichiometry) by the yield percentage:
actual moles of chloromethane = moles of chloromethane * yield percentage (as a decimal)
Finally, we can calculate the mass of chloromethane formed by multiplying the actual moles by the molar mass:
mass of chloromethane formed = actual moles of chloromethane * molar mass of chloromethane
Substituting the values into the equation, we can find the mass of chloromethane formed.
To find the mass of chloromethane (CH3Cl) formed, we need to determine the limiting reactant. This reactant will be completely consumed in the reaction, and its amount will determine the maximum amount of product formed.
First, we need to calculate the number of moles of each reactant:
The molar mass of methane (CH4) is:
12.01 g/mol (C) + 1.01 g/mol (H) × 4 = 16.04 g/mol
The number of moles of methane is:
20.5 g / 16.04 g/mol ≈ 1.28 mol
The molar mass of chlorine (Cl2) is:
35.45 g/mol × 2 = 70.90 g/mol
The number of moles of chlorine is:
45.0 g / 70.90 g/mol ≈ 0.63 mol
Next, we can determine the stoichiometric ratio between methane and chloromethane from the balanced equation:
CH4 + Cl2 → CH3Cl + HCl
From the balanced equation, we know that 1 mol of methane reacts with 1 mol of chlorine to produce 1 mol of chloromethane.
Since both reactants have a 1:1 mole ratio with the product, the limiting reactant will be the one with fewer moles. In this case, chlorine is the limiting reactant, as it has fewer moles (0.63 mol) compared to methane (1.28 mol).
Now, we can calculate the theoretical yield of chloromethane using the limiting reactant:
The molar mass of chloromethane (CH3Cl) is:
12.01 g/mol (C) + 1.01 g/mol (H) × 3 + 35.45 g/mol (Cl) = 50.49 g/mol
The theoretical yield of chloromethane is:
0.63 mol × 50.49 g/mol ≈ 31.82 g
Finally, we can calculate the actual yield of chloromethane using the given 75.0% yield:
The actual yield of chloromethane is:
75.0% × 31.82 g ≈ 23.87 g
Therefore, the mass of chloromethane (CH3Cl) formed is approximately 23.87 g.