# How to calculate volume of gas?

MgCO3 --> MgO + CO2

What volume of CO2 (in dm^3) would have been given off when measured at room temperature and pressure?

MgCO3 = 1.47 g

Molar mass of MgCO3 = 84.3 g mol^-1

Please help! I'm really confused...

### 2 Answers

- Trevor HLv 77 years agoFavorite Answer
From the equation :

1 mol MgCO3 will produce 1 mol CO2

mol MgCO3 = 1.47/84.3 = 0.0174 mol MgCO3

This will produce 0.0174 mol CO2

I am always at a loss as to what is meant by "room temperature and pressure". I referred to Wikipedia:

An unofficial, but commonly used standard is standard ambient temperature and pressure (SATP) as a temperature of 298.15 K (25 °C, 77 °F) and an absolute pressure of 100 kPa (14.504 psi, 0.987 atm).

Use the gas equation to calculate volume using :

P = 0.987 atm

V = ???

n = mol CO2 = 0.0174

R = 0.082057

T = 298.15

Substitute:

0.987 * V = 0.0174 * 0.082057*298.15

V = 0.4266 / 0.987

V = 0.432 L

Volume of CO2 produced = 0.432L

- BobLv 47 years ago
1.47 g MgCO3 * (1 mol MgCO3 / 84.3 g MgCO3) = .0174 mol MgCO3

1 to 1 mol ratio

mol MgCO3 = mol CO2

mol CO2 = .0174

.0174 mol CO2 * (22.4 L / 1 mol CO2) = .390 L CO2

1 L = 1 dm^3

.390 dm ^2 = CO2