Question

Constants: A: MW = 150 g/mol; B: = MW 100 g/mol; C: MW = 200 g/mol. 2.0 g C was made from 4.5 g A and 4.0 g

Answer 1

Answer:

a. 100%

b. 133%

c. 300%

Explanation:

To find yield first we need to determine theoretical yield converting each reactant to moles and find limitng reactant for each reaction:

Moles A:

4.5g * (1mol / 150g) = 0.03 moles

Moles B:

4.0g * (1mol / 100g) = 0.04 moles

a. For a complete reaction of 0.03 moles of A are needed:

0.03 moles A * (1 mole B / 3 moles A) = 0.01 moles of B

As there are 0.04 moles of B, A is limiting reactant.

Theoretical moles and mass of C are:

0.03 moles A * (1 mole C / 3 moles A) = 0.01 moles of C.

0.01 moles of C * (200g / mol) = 2g are produced.

Yield is:

2g / 2g * 100 = 100%

b. For a complete reaction of 0.03 moles of A are needed:

0.03 moles A * (3 mole B / 2 moles A) = 0.045 moles of B

As there are 0.04 moles of B, B is limiting reactant.

Theoretical moles and mass of C are:

0.04 moles B * (1 mole C / 3 moles B) = 0.0133 moles of C.

0.0133 moles of C * (200g / mol) = 2.67g are produced.

Yield is:

2.67g / 2g * 100 = 133%

c. For a complete reaction of 0.03 moles of A are needed:

0.03 moles A * (1 mole B / 1 moles A) = 0.03 moles of B

As there are 0.04 moles of B, A is limiting reactant.

Theoretical moles and mass of C are:

0.03 moles A * (1 mole C / 1 moles A) = 0.03 moles of C.

0.03 moles of C * (200g / mol) = 6g are produced.

Yield is:

6g / 2g * 100 = 300%