The volume equivalent of CO2(at stp) in the reaction.
NaHCO3+HCL=NaCL+H20+CO2 is
To determine the volume equivalent of CO2 produced in the reaction of NaHCO3 + HCl = NaCl + H2O + CO2, you need to follow these steps:
1. Write and balance the chemical equation:
NaHCO3 + HCl -> NaCl + H2O + CO2
2. Convert the given reactants or products to moles (if the units given are not already moles). To do this, you need to know the molar masses of the substances involved:
Molar mass of NaHCO3 = 22.99 + 1.01 + 12.01 + (3 × 16.00) = 84.01 g/mol
Molar mass of HCl = 1.01 + 35.45 = 36.46 g/mol
3. Determine the limiting reactant. This is the reactant that will be completely consumed in the reaction. To do this, you compare the number of moles of each reactant to their stoichiometric coefficients (the numbers in front of each substance in the balanced equation).
For example, if you have 2 moles of NaHCO3 and 4 moles of HCl, the limiting reactant would be NaHCO3 since it has a stoichiometric coefficient of 1 and HCl has a coefficient of 1 as well.
4. Use the stoichiometry of the balanced equation to calculate the moles of CO2 produced. The stoichiometric coefficient of CO2 is also 1.
So, if NaHCO3 is the limiting reactant, 1 mole of NaHCO3 will produce 1 mole of CO2.
5. Convert the moles of CO2 to volume at STP (standard temperature and pressure), which is 0 degrees Celsius and 1 atmosphere pressure.
1 mole of any gas at STP has a volume of 22.4 liters.
Therefore, the volume equivalent of CO2 produced in the reaction NaHCO3 + HCl = NaCl + H2O + CO2, when NaHCO3 is the limiting reactant, is 22.4 liters.