posted by Fai on .
A 0.4671g sample containing sodium bicarbonate was titrated with HCL requiring 40.72ml. The acid was standardized by titrating 0.1876g of sodium carbonate FW 106mg/mmol requiring 37.86ml of the acid. Find the percentage of NAHO3 FW 84mg/mmol in the sample
Who helps me to solve it. This question is too difficult for me. Pls help
First determine the molarity of the HCl from the standardization part of the problem (the second part of the problem) using 0.1876 g NaHCO3 and 37.86 mL HCl.
NaHCO3 + HCl ==> NaCl + H2O + CO2
mols NaHCO3 = grams/molar mass
mols HCl = mols NaHCO3
M HCl = mols HCl/L HCl.
Then determine percent NaHCO3 in the unknown which is the first part.
mols HCl = M x L = ?M HCl from above x 0.04072L = ?
mols NaHCO3 = mols HCl
g NaHCO3 = mols x molar mass
%NaHCO3 = (grams NaHCO3/0.4674)*100 = ?
How many grams of NaCl would be formed from 100 grams of Na2CO3 in the following reaction?
Na2CO3 + 2HCl→2 NaCl+CO2+H2O