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    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?
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    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 = \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)
 
 
 
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