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

The volume of benzene required is 42.5 mL of benzene.

The equation of the reaction is;  2C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) ΔH°rxn = -6278 kJ

If 2 mole of benzene produces -6278 kJ of heat

x moles of benzene produces 1.5 x 103 kJ

x =  2 mole ×  -1.5 x 10^3 kJ/ -6278 kJ

x = 0.48 moles of benzene

The mass of benzene required = number of moles of benzene x molar mass of benzene

Molar mass of benzene = 78 g/mol

Mass of benzene required = 0.48 moles of benzene × 78 g/mol

= 37.44 g

Density of benzene = 0.88 g/mL

But; density = mass/volume

Volume = mass/density

volume = 37.44 g/0.88 g/mL

volume = 42.5 mL of benzene

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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