Na2CO3+ 2HCl -------> 2NaCl+ CO2+ H2O
0.05.............0.1................0.1......0.05.....0.05
a)nNa2CO3=\(\dfrac{53\cdot10\%}{106}=0.05mol\)
a)=>mNaCl=0.1*58.5=5.85 g
b) mdd=mNa2CO3+mHCl-mCO2=53+(0.1*36.5/20%)-0.05*44=69.05 g
=>C%NaCl =(5.85*100)/69.05= 8.47%