If you used 2.0mL of 1.0M H2SO4 and 3.0g of NaHCO3 how many moles of CO2 would be produced?
This is a limiting reagent (LR) problem and we know that because amounts are given for BOTH reactants.
H2SO4 + 2NaHCO3 ==> Na2SO4 + 2H2O + 2CO2
mols H2SO4 = M x L = 0.002
mols NaHCO3 = g/molar mass = approx 0.036 but that's only an estimate and you should do all of these calculations over.
Now convert each to mols of the product.
First H2SO4:
0.002 mols H2SO4 x (2 mols CO2/1 mol H2SO4) = 0.004 mols CO2.
Then NaHCO3:
0.036 mols NaHCO3 x (2 mols CO2/2 mols NaHCO3) = 0.036 x (2/2) = 0.036 mols CO3.
The two numbers don't agree (they rarely do in LR problems) so one of them must be wrong; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR. So we will produce 0.005 mols CO2
To determine the number of moles of CO2 produced, we need to use stoichiometry, which involves balancing the chemical equation and using the mole ratios.
First, we need to write the balanced chemical equation for the reaction between H2SO4 (sulfuric acid) and NaHCO3 (sodium bicarbonate):
H2SO4 + 2 NaHCO3 → Na2SO4 + 2 CO2 + 2 H2O
In this equation, we can see that 2 moles of NaHCO3 produce 2 moles of CO2.
Given:
Volume of H2SO4 = 2.0 mL = 0.002 L (since 1 mL = 0.001 L)
Concentration of H2SO4 = 1.0M
Mass of NaHCO3 = 3.0 g
Molar mass of NaHCO3 = 22.99 g/mol + 1.01 g/mol + 12.01 g/mol + 3 * 16.00 g/mol = 84.01 g/mol
Now, let's calculate the number of moles of NaHCO3:
Number of moles = mass / molar mass
Number of moles of NaHCO3 = 3.0 g / 84.01 g/mol ≈ 0.0357 mol
Since the balanced equation shows that 2 moles of NaHCO3 produce 2 moles of CO2, we can conclude that 0.0357 mol of NaHCO3 will produce 0.0357 mol of CO2.
Therefore, 0.0357 mol of CO2 would be produced from this reaction.