4.3 moles of a gas are at a temperature of 28 degrees * C with a pressure of 1.631 atm. What volume does the gas occupy?

65.2L

Explanation:

Using the general gas equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (Litres)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (Kelvin)

According to the information provided in this question,

P = 1.631 atm

V = ?

n = 4.3 moles

T = 28°C = 28 + 273 = 301K

Using PV = nRT

V = nRT/P

V = 4.3 × 0.0821 × 301 ÷ 1.631

V = 106.26 ÷ 1.631

V = 65.15

Volume of the gas = 65.2L

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it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2