Poisonous carbon monoxide gas is a product of incomplete burning of a carbon containing fuel. What is the volume of 1.63 moles of CO gas at 11 °C and 600 mm Hg?

Please show work!! I am having trouble with this topic ngl

48.11L

Explanation:

The equation below shows how limited supply of air results in incomplete combustion of carbon which then leads to the formation of carbon monoxide gas

2C(s) + O2(g) ---> 2CO(g)

Given that:

Volume of CO (V) = ?

Temperature of CO gas (T) = 11°C

[Convert 11°C to Kelvin by adding 273

11°C + 273 = 284K]

Pressure of CO (P) = 600 mmHg

[convert 600 mmHg to atmosphere

If 760 mmHg = 1 atm

600 mmHg = 600/760 = 0.789 atm]

Number of moles of CO gas (n) = 1.63 moles

Molar gas constant (R) is a constant with a value of 0.082 atm L K-1 mol-1

Then, apply the formula for ideal gas equation

pV = nRT

0.789 atm x V = 1.63 moles x 0.082 atm L K-1 mol-1 x 284K

0.789 atm•V = 37.96 atm•L

Divide both sides by 0.789 atm to get V

0.789 atm•V/0.789 atm = 37.96 atm•L/0.789 atm

V = 48.11L

Thus, the volume of CO gas is 48.11 litres

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