Balance the eqn
Na2CO3 + 2HCl -> 2NaCl +H20 + CO2
From the balanced eqn you can see that one mole of CO2 is produced.
Then at stp 1 mole of gas takes up 24 litres (I think, check this on google) therefore v= 24 litres.
:)
Calculate the volume of Carbon dioxide gas generated when 40.0 g of sodium carbonate reacts with hydrocholric acid according to the following reaction. The gas is to be collected at STP
Na2CO3 + HCL-----> NaCL +H2O + CO2(g)
2 answers
Convert 40 g Na2CO3 to moles. moles = grams/molar mass. Since 1 mole Na2CO3 produces 1 mole CO2, moles Na2CO3 above will contribute that many moles CO2 @ STP. Remember 1 mole CO2 @ STP occupies 22.4 L.
2 answers