We have 10.0 g of sucrose (C₁₂H₂₂O₁₁) and we want to study the combustion of this compound.

1) How many molecules do you have in 10.0 g of sucrose?

2) What is the chemical equation for the combustion of sucrose?

3) We have 12.0 g of oxygen. What is the limiting reagent of this reaction?

Question

We have 10.0 g of sucrose (C₁₂H₂₂O₁₁) and we want to study the combustion of this compound.

1) How many molecules do you have in 10.0 g of sucrose?

2) What is the chemical equation for the combustion of sucrose?

3) We have 12.0 g of oxygen. What is the limiting reagent of this reaction?

| General Chemistry 2

Accepted Answer

1) n = $\frac{N}{N_{A}}$ and n = $\frac{m}{M}$ with M = 12 M_C + 22 M_H + 11 M_O = 12 x 12.0 + 22 x 1.01 + 11 x 16.0 = 342 g.mol^-1

⇒ N = n x N_A = $\frac{m}{M}$ x N_A

⇒ N = $\frac{10.0}{342}$ x 6.022 x 10²³ = 1.76 x 10²² molecules of sucrose

2) C₁₂H₂₂O₁₁ (l) + 12 O₂ (g) → 12 CO₂ (g) + 11 H₂O (l)

3) n_sucrose = $\frac{m_{sucrose}}{M_{sucrose}}$ = $\frac{10.0}{342}$ = 2.92 x 10^-2 mol

Moles of O₂ needed to completely react with 10.0 g of sucrose:

12 moles of O₂ react with 1 mole of sucrose

⇒ n_{O2 needed} = 12 x n_sucrose = 3.51 x 10^-1 mol

n_O2 = $\frac{m_{O 2}}{M_{O 2}}$ = $\frac{12.0}{32}$ = 3.75 x 10^-1 mol > n_{O2 needed}

The sucrose is the limiting reagent of this reaction

Midterm 1 - Question 4 | General Chemistry 2

General Chemistry 2 - Midterm 1