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

the answer is a and have a great day

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.