Question

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?

Answer 1

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.

Answer 2

what is the balanced equation?

how do i convert moles of CH4 and Cl2 to moles of the product??

Answer 3

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 :)

Answer 4

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.