Question

Limiting Reactant Calculations C6H1206(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g) Starting with 25.0 g C6H12O6 and 15.5 g 02, what is the theoretical yield of CO2 in grams? Calculation based upon C6H1206 Calculation based upon 02 What is the limiting reactant? What is the theoretical yield?

Answer 1

1)

Molar mass of C6H12O6,
MM = 6*MM(C) + 12*MM(H) + 6*MM(O)
= 6*12.01 + 12*1.008 + 6*16.0
= 180.156 g/mol

mass of C6H12O6 = 25 g
mol of C6H12O6 = (mass)/(molar mass)
= 25/1.802*10^2
= 0.1388 mol

According to balanced equation
mol of CO2 formed = (6/1)* moles of C6H12O6
= (6/1)*0.1388
= 0.8326 mol


Molar mass of CO2,
MM = 1*MM(C) + 2*MM(O)
= 1*12.01 + 2*16.0
= 44.01 g/mol


mass of CO2 = number of mol * molar mass
= 0.8326*44.01
= 36.64 g
Answer: 36.6 g

2)

Molar mass of O2 = 32 g/mol

mass of O2 = 15.5 g
mol of O2 = (mass)/(molar mass)
= 15.5/32
= 0.4844 mol

According to balanced equation
mol of CO2 formed = moles of O2
= 0.4844 mol


Molar mass of CO2,
MM = 1*MM(C) + 2*MM(O)
= 1*12.01 + 2*16.0
= 44.01 g/mol


mass of CO2 = number of mol * molar mass
= 0.4844*44.01
= 21.32 g
Answer: 21.3 g

3)
Mass of CO2 is less in case of O2
So, O2 is limiting
Answer: O2

4)
Answer: 21.3 g