When 20.5g of methane and 45.0g of chlorine gas undergo a reaction that has a 75.0% yeild, what mass of chloromethane forms?
STEP 1 equation:
CH4 + Cl2 --> CH3Cl + HCl
STEP 2 moles of CH4:
CH4 = 12.01 g + (1.008 g x 4) = 16.042 g/mol
20.5 g CH4 x (1 mol CH4 / 16.042 g CH4) = 1.28 mol CH4
STEP 3 moles of Cl2:
Cl2 = (35.45 g x 2) = 70.9 g g/mol
45.0 g Cl2 x (1 mol Cl2 / 70.9 g Cl2) = 0.63 mol Cl2
STEP 4 limiting reagent:
0.63 mol Cl2 < 1.28 mol CH4
Therefore, Cl2 is the limiting reagent because it is the smaller number.
STEP 5 mol of CH3Cl:
USE THE LIMITING REAGENT Cl2 to find the ratio...
1 Cl2 : 1 CH3Cl (1:1 ratio)
Cl2 = 0.63 mol therefore CH3Cl = 0.63 mol
STEP 6 g of CH3Cl:
CH3Cl = 12.01 g + (1.008 g x 3) + 35.44 g = 50.484 g/mol
0.63 mol CH3Cl x (50.484 g CH3Cl / 1 mol CH3Cl) = 31.8 g CH3Cl
STEP 7 75% of CH3Cl:
31.8 g of CH3Cl x .75 = 24 g CH3Cl <<< FINAL ANSWER
Hope that helps :)
This is a limiting reagent problem.
Do you have an equation? Write it and balance.
Convert 20.5 g CH4 to moles. moles = grams/molar mass.
Convert 45.0 g Cl2 to moles.
Using the coefficients in the balanced equation, convert moles CH4 to moles of the product.
Do the same for moles Cl2.
This will give you two numbers for the product; obviously both can't be correct. The smaller number of moles of the product is ALWAYS the correct number to choose. The reactant that provides that small number is the limiting reagent.
Finally, convert moles of the product to grams. g = moles x molar mass.
Post your work if you get stuck.
Write the balanced redox reaction and Calculate the potential for the titration of 50.0 mL of 0.0500 M Sn2+ after adding 10, 25 and 40 mL of 0.100 M Tl3+. Both the analyte and the titrant are 1.0 M in HCl. E0 Tl4+/Tl3+ = +0.77 V and E0 Sn2+/Sn4+ = +0.139 V.
Well, let's calculate the mass of chloromethane formed during this reaction. Chloromethane, also known as methyl chloride, has the chemical formula CH3Cl.
First, we need to determine the limiting reactant. To do this, we can convert both methane and chlorine gas to moles using their molar masses. The molar mass of methane (CH4) is approximately 16.04 g/mol, and the molar mass of chlorine (Cl2) is about 70.90 g/mol.
For methane:
Moles of CH4 = mass of CH4 / molar mass of CH4
Moles of CH4 = 20.5 g / 16.04 g/mol
For chlorine gas:
Moles of Cl2 = mass of Cl2 / molar mass of Cl2
Moles of Cl2 = 45.0 g / 70.90 g/mol
Now, let's compare the moles of each reactant to see which one is the limiting reactant. The reaction equation is:
CH4 + Cl2 → CH3Cl + HCl
The stoichiometric ratio of methane to chloromethane is 1:1. Therefore, we need equal moles of methane and chlorine to react completely. Whichever reactant has fewer moles will be the limiting reactant.
Now, let's calculate the moles of each reactant:
Moles of CH4 = 20.5 g / 16.04 g/mol ≈ 1.28 mol
Moles of Cl2 = 45.0 g / 70.90 g/mol ≈ 0.63 mol
Since we need the same number of moles of both reactants, chlorine gas is the limiting reactant because it has fewer moles.
Now, let's calculate the moles of chloromethane formed based on the limiting reactant (Cl2). The stoichiometric ratio of Cl2 to CH3Cl is 1:1.
Moles of CH3Cl = Moles of Cl2 = 0.63 mol
Finally, let's convert the moles of chloromethane to grams using its molar mass (50.49 g/mol):
Mass of CH3Cl formed = Moles of CH3Cl × Molar mass of CH3Cl
Mass of CH3Cl formed = 0.63 mol × 50.49 g/mol
So, the mass of chloromethane formed is approximately 31.85 g.
Remember, this is based on the assumption of a 100% yield. Since you mentioned that the yield is 75.0%, the actual mass of chloromethane formed would be 75.0% of 31.85 g.
But hey, at least we've got math jokes to lighten up the chemistry! Why did the mole go undercover? Because it wanted to be an "under-covert" agent!
To find the mass of chloromethane that forms in the reaction, we need to determine the limiting reactant first.
1. Start by writing the balanced chemical equation for the reaction:
CH4 (methane) + Cl2 (chlorine gas) → CH3Cl (chloromethane) + HCl (hydrogen chloride)
2. Calculate the number of moles for each reactant:
Number of moles of methane = mass of methane / molar mass of methane
Number of moles of chlorine gas = mass of chlorine gas / molar mass of chlorine gas
The molar mass of methane (CH4) is approximately 16 g/mol (1 carbon atom with a molar mass of 12 g/mol and 4 hydrogen atoms with a molar mass of 1 g/mol each).
The molar mass of chlorine gas (Cl2) is approximately 71 g/mol (2 chlorine atoms with a molar mass of 35.5 g/mol each).
3. Use the coefficients in the balanced equation to determine the moles of chloromethane formed:
According to the balanced equation, the ratio of methane to chloromethane is 1:1. Thus, the number of moles of chloromethane formed is equal to the number of moles of methane.
4. Determine the limiting reactant:
The limiting reactant is the one that produces the least amount of product. To find it, compare the moles of methane and chlorine gas calculated in step 2. If the ratio of their coefficients in the balanced equation is different from 1:1, divide the moles by the respective coefficients to compare them.
5. Calculate the theoretical yield:
The theoretical yield is the maximum amount of product that can be formed from the limiting reactant. To calculate it, multiply the number of moles of the limiting reactant (in this case, methane) by the molar mass of chloromethane.
6. Determine the actual yield:
The actual yield is given in the problem statement as a percentage of the theoretical yield. Multiply the theoretical yield by the given percentage (0.75 in this case).
7. Calculate the mass of chloromethane formed:
Multiply the actual yield by the molar mass of chloromethane to obtain the mass of chloromethane produced.
By following these steps, you can find the mass of chloromethane that forms in the given reaction.
what is the balanced equation?
how do i convert moles of CH4 and Cl2 to moles of the product??