4.20g of NaHCO3 are reacted with 40cm^3(cubed) of HCl of concentration 2.00 mol dm^-3(cubed). The equation for this reaction is:
NaHCO3 + HCl -------> NaCl + CO2 + H2O
(1 mole of any gas occupies 24.0dm^3(cubed) at room temperature and pressure)
Questions:
1) How many moles of NaHCO3 were used? - i got the answer 0.05 mol......is this correct?
2) How many moles of HCl are present in 40.0cm^3(cubed) of 2.00 mol dm^-3(cubed) solution? the answer i got is 0.08 mol.....is this correct?
3) How many moles of HCl are left unreacted? my answer is 1.87 mol.....is this correct?
4) How many moles of CO2 are formed? i wasn't sure how to do this.
5) What volume of CO2 is formed? i wasn't sure how to do this.
NaHCO3 + HCl -------> NaCl + CO2 + H2O
(1 mole of any gas occupies 24.0dm^3(cubed) at room temperature and pressure)
Questions:
1) How many moles of NaHCO3 were used? - i got the answer 0.05 mol......is this correct?
2) How many moles of HCl are present in 40.0cm^3(cubed) of 2.00 mol dm^-3(cubed) solution? the answer i got is 0.08 mol.....is this correct?
3) How many moles of HCl are left unreacted? my answer is 1.87 mol.....is this correct?
4) How many moles of CO2 are formed? i wasn't sure how to do this.
5) What volume of CO2 is formed? i wasn't sure how to do this.