Question

given the following equation:
2 kclo₃ → 2 kcl + 3 o₂
how many moles of o₂ are produced when 50 mol of kclo₃ reacts?
mol o₂
question 7
given the following equation:
2 c₄h₁₀ + 13 o₂ → 8 co₂ + 10 h₂o
how many grams of c₄h₁₀ are needed to produce 3.1 mol of co₂?
g c₄h₁₀

Explanation:

Question 6 (Moles of \( O_2 \) from \( KClO_3 \) reaction)

Step1: Identify mole ratio

From the equation \( 2\ KClO_3
ightarrow 2\ KCl + 3\ O_2 \), the mole ratio of \( KClO_3 \) to \( O_2 \) is \( 2:3 \).

Step2: Set up proportion

Let \( x \) be moles of \( O_2 \). Then \( \frac{2\ mol\ KClO_3}{3\ mol\ O_2} = \frac{50\ mol\ KClO_3}{x\ mol\ O_2} \).

Step3: Solve for \( x \)

Cross - multiply: \( 2x = 3\times50 \). So \( 2x = 150 \), and \( x=\frac{150}{2}=75 \).

Step1: Identify mole ratio

From \( 2\ C_4H_{10}+13\ O_2
ightarrow8\ CO_2 + 10\ H_2O \), mole ratio of \( C_4H_{10} \) to \( CO_2 \) is \( 2:8 = 1:4 \).

Step2: Find moles of \( C_4H_{10} \)

Let \( y \) be moles of \( C_4H_{10} \). \( \frac{1}{4}=\frac{y}{3.1\ mol\ CO_2} \), so \( y=\frac{3.1}{4}=0.775\ mol\ C_4H_{10} \).

Step3: Calculate molar mass of \( C_4H_{10} \)

Molar mass of \( C_4H_{10} \): \( 4\times12.01+10\times1.008 = 48.04 + 10.08=58.12\ g/mol \).

Step4: Find mass of \( C_4H_{10} \)

Mass = moles×molar mass = \( 0.775\ mol\times58.12\ g/mol\approx45.04\ g \).

Answer:

75

Question 7 (Grams of \( C_4H_{10} \) needed for \( CO_2 \) production)