Stoichiometric Calculations: Write the complete solutions
Exactly 0.2220 g of pure Na2CO3 was dissolved in 100.0 ml of 0.0731 MHCI. What was the molar concentration of the excess reactant (HCI or Na2CO3)
First, let's find the number of moles of Na2CO3 that was dissolved:
Molar mass of Na2CO3 = 2(22.98977) + 12.0107 + 3(15.9994) = 105.9888 g/mol
Number of moles = Mass / Molar mass = 0.2220 g / 105.9888 g/mol = 0.002094 mol
Since the reaction between Na2CO3 and HCl is:
Na2CO3 + 2HCl -> 2NaCl + H2O + CO2
We can see that 1 mole of Na2CO3 reacts with 2 moles of HCl. Therefore, the number of moles of HCl needed would be:
0.002094 mol Na2CO3 * (2 mol HCl / 1 mol Na2CO3) = 0.004188 mol HCl
Now, we know that the initial concentration of HCl is 0.0731 M, and the volume of the HCl solution is 100.0 ml = 0.100 L. Using the formula for molarity:
M1V1 = M2V2
(0.0731 M) * (0.100 L) = M2 * (0.100 L + 0.100 L)
0.00731 mol = 0.2 M * (0.200 L)
M2 = 0.00731 mol / 0.200 L = 0.03655 M
Therefore, the molar concentration of the excess reactant (HCl) is 0.03655 M.