CH4+Cl2->CH2Cl2+HCl.
If 6.00 g of CH4 are mixed with 6.00 g of CL2, how much HCl can be made? What is the percent yield if 2.5 g are recovered from the reaction in the lab?
Here is a site. A lot of reading but the essentials are here. The problem you posted is a limiting reagent problem. If you have trouble, reply with details about what you don't understand.
http://www.shodor.org/UNChem/basic/stoic/
To determine how much HCl can be made, we need to first find the limiting reactant. The limiting reactant is the reactant that limits the amount of product that can be formed.
To find the limiting reactant, we compare the mole ratios of the reactants to the ratio in the balanced chemical equation. The balanced chemical equation for the reaction is:
CH4 + Cl2 -> CH2Cl2 + HCl
The mole ratio of CH4 to HCl is 1:1, and the mole ratio of Cl2 to HCl is 1:2.
First, we need to convert the mass of CH4 and Cl2 into moles. The molar mass of CH4 is 16.04 g/mol and the molar mass of Cl2 is 70.91 g/mol.
Moles of CH4 = 6.00 g CH4 / 16.04 g/mol = 0.374 mol CH4
Moles of Cl2 = 6.00 g Cl2 / 70.91 g/mol = 0.085 mol Cl2
Next, we compare the mole ratios of CH4 and Cl2 to HCl. Since the mole ratio of CH4 to HCl is 1:1, and the moles of CH4 is the same as moles of HCl, we conclude that both reactants are in a 1:1 ratio. Therefore, the limiting reactant is CH4.
Now, we need to calculate the moles of HCl that can be formed based on the moles of CH4:
Moles of HCl = 0.374 mol CH4
Finally, we can calculate the mass of HCl formed based on the moles of HCl:
Mass of HCl = Moles of HCl * molar mass of HCl = 0.374 mol * 36.46 g/mol = 13.64 g
Therefore, 13.64 g of HCl can be made.
To calculate the percent yield, we need to compare the actual yield to the theoretical yield. The actual yield is given as 2.5 g of HCl.
Percent yield = (actual yield / theoretical yield) * 100
Percent yield = (2.5 g / 13.64 g) * 100 = 18.31%
Therefore, the percent yield is 18.31%.