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.
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)
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