Question

What volume of benzene (C6H6, d= 0.88 g/mL, molar mass = 78.11 g/mol) is required to produce 1.5 x 103 kJ of heat according to the following reaction?2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) ΔH°rxn = -6278 kJ

Answer 1

Answer:

We need 42.4 mL of benzene to produce 1.5 *10³ kJ of heat

Explanation:

Step 1: Data given

Density of benzene = 0.88 g/mL

Molar mass of benzene = 78.11 g/mol

Heat produced = 1.5 * 10³ kJ

Step 2: The balanced equation

2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g)         ΔH°rxn = -6278 kJ

Step 3: Calculate moles of benzen

1.5 * 10³ kJ * (2 mol C6H6 / 6278 kJ) = 0.478 mol C6H6

Step 4: Calculate mass of benzene

Mass benzene : moles benzene * molar mass benzene

Mass benzene= 0.478 * 78.11 g

Mass of benzene = 37.34 grams

Step 5: Calculate volume of benzene

Volume benzene = mass / density

Volume benzene = 37.34 grams / 0.88g/mL

Volume benzene = 42.4 mL

We need 42.4 mL of benzene to produce 1.5 *10³ kJ of heat