Answer:
1.0979g of Na2CO3
0.8471g of NaHCO3
Explanation:
NaHCO3 and Na2CO3 reacts with HCl as follows:
NaHCO3 + HCl → NaCl + H2O + CO2
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
Moles of HCl required were:
0.03531L * (0.8724mol / L) = 0.03080 moles of HCl
With this information we can write 2 equations:
1.9450g = X + Y (1)
X is the mass of NaHCO3 and Y is mass of Na2CO3
Also we can write:
0.03080 moles = X/84 + 2Y/106 (2)
Where 84 is molar mass of NaHCO3 and 106 is molar mass of Na2CO3, 2 is due the reaction (2 moles of HCl react per mole of Na2CO3)
Replacing (1) in (2):
0.03080 moles = (1.9450-Y)/84 + 2Y/106
0.03080 = 0.023155 - Y/84 + Y/53
0.007645 = 0.00696316Y
Y = 1.0979g of Na2CO3
And mass of NaHCO3 =
1.9450g - 1.0979g =
0.8471g of NaHCO3
