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:
<em>Moles A:</em>
4.5g * (1mol / 150g) = 0.03 moles
<em>Moles B:</em>
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%
