How many grams of dichloromethane are formed from 1500.0 grams of methane, it the percent yield for the reaction is 82.5%?
You don't have an equation to use; is this it?
CH4 + 2Cl2 ==> CH2Cl2 + 2HCl
mols CH4 = grams/molar mass = 1500/16 = approx 100 but you need to do it more accurately.
Using the coefficients in the balanced equation, convert mols CH4 to mols of Ch2Cl2. That will be approx 100 x (1 mol CH2Cl2/1 mol CH4) = approx 100 mols CH2Cl2.
Grams CH2Cl2 = approx 100 mols x molar mass = 100 x approx 81 = approx 8100 grams if the reaction were 100%. This is the theoretical yield (TY). You want to find the actualyield (AY)
%yield = [AY/TY]*100
You know the %yield and the TY, solve for (AY).
To find the number of grams of dichloromethane formed, we need to determine the actual yield of the reaction, given the percent yield.
Percent yield is the ratio of the actual yield to the theoretical yield, expressed as a percentage. In this case, the percent yield is given as 82.5%.
The theoretical yield is the maximum amount of product that could be formed based on the balanced chemical equation. In this reaction, methane (CH4) reacts with chlorine gas (Cl2) to produce dichloromethane (CH2Cl2).
The balanced chemical equation for the reaction is:
CH4 + Cl2 → CH2Cl2 + HCl
From the equation, we can see that the stoichiometry between methane and dichloromethane is 1:1. This means that for every 1 mole of methane, 1 mole of dichloromethane is formed.
To find the moles of methane, we divide the given mass (1500.0 grams) by the molar mass of methane (16.04 g/mol):
moles of methane = 1500.0 g / 16.04 g/mol ≈ 93.58 moles
Since the stoichiometry of methane to dichloromethane is 1:1, the number of moles of dichloromethane formed will be the same as the moles of methane.
Now, to find the actual yield of dichloromethane, we multiply the moles of dichloromethane by its molar mass (84.93 g/mol):
actual yield = 93.58 moles × 84.93 g/mol ≈ 7949.27 grams
Finally, to find the grams of dichloromethane formed based on the percent yield, we multiply the actual yield by the percent yield:
grams of dichloromethane formed = actual yield × percent yield / 100
grams of dichloromethane formed = 7949.27 g × 82.5 / 100 ≈ 6559.14 grams
Therefore, approximately 6559.14 grams of dichloromethane are formed from 1500.0 grams of methane, given a percent yield of 82.5%.