Convert 57.0 L carbon monoxide (at STP) to grams.

Answer 1

PV=nRT

At STP, T = 273.15K; P = 1 atm

R is constant and equals 0.0821 (atm L)/(mol K)

n = PV/RT

n = (1 atm x 57 L) / ([0.0821 (atm L)/(mol K)] / 273.15 K)

n = 2.5417 moles

mass = molar mass x moles

molar mass Carbon Monoxide = 12.01 g per mole + 16 g per mole = 28.01 g per mole

mass = (28.01 g per mole) x 2.5417 mol

mass = 71.17 g

Answer 2

When conditions are at STP (Standard Temperature and Pressure), you should memorize the value

1 mol of gas = 22.4 liters. Using this conversion, you can avoid using the ideal gas law of PV = nRT

In the present problem we will use this value to answer the question.

57.0 L x 1 mol / 22.4 L = 2.54 moles CO

Molar mass CO = 28 g / mol

grams CO = 2.54 mols CO x 28 g / mol = 71.1 g CO