The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Calculating volume of gas

Can anyone help me with this question?

4.20g of NaHCO3 are reacted with 40.0cm3 of HCl of concentration 2.00 mol dm-3. The acid is in excess. The equation for the reaction is:

NaHCO3 + HCl = NaCl + H2O + CO2

What volume of CO2 is formed?
First you need to work out the number of moles of NaHCO3 in 4.20g - the molar mass is about 84gmol-1. This gives 4.20g/84gmol-1=0.050mol

Given that the acid is in excess, we know that all of the NaHCO3 will react. Looking at the ratio of NaHCO3 to CO2 in the equation, they are 1:1, so for every mol of NaHCO3 we will produce the same amount of CO2. This means 0.050mol of CO2 are produced.

To calculate the volume, we would normally use pV=nRT. We don't know any of the conditions (P or T) of the gas, so we must assume it's at standard temperature and pressure/STP (roughly 273K and 100000Pa). With these values, we could either use the ideal gas equation, or the fact that 1 mol occupies 22.4dm3 at STP. Personally I prefer to use the gas equation so that I don't have to remember that number, but I see a lot of people prefer to use that.

Either way, you would get about 1.12dm3. (either by doing 0.05mol x 22.4m3mol-1 or using V = nRT/p = 0.05mol×8.31JK−1mol−1×273K100000Pa\frac{0.05mol \times 8.31JK^{-1}mol^{-1} \times 273K}{100000Pa}. If you use the gas equation remember your answer will be in m3 rather than dm3)

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you