Question

Consider the reaction:
2AT + 2HO + ME -------> JO + LL + IB + EE
MM 101.1 12.01 32.07 28.01 44.01 28.02 158.3

a. If 40.00 g of AT, 30.00 g of HO, and 30.00 g of ME are allowed to react, which reactant is the limiting?
b. How many grams of EE can be produced?
c. How much of EE (in moles) must be produced in order to have 85% completion?

Answer 1

a. n(AT)=m/M=40g/101.1g/mol=0.4mol
n(HO)=30g/12.01g/mol=2.5mol
n(ME)=30g/32.07g/mol=0.9mol
AT is the limiting reactant
b. n(EE)=n(AT)/2=0.4/2=0.2mol
m(EE)=n×M=0.2mol×158.3g/mol=31.66g
c. n(EE)=0.2×0.85=0.17mol