Đáp án:
$5,98\%$
Giải thích các bước giải:
$a)Na_2CO_3+2HCl \to 2NaCl+CO_2+H_2O\\b)m_{Na_2CO_3}=200.10,6\%=21,2(g)\\n_{Na_2CO_3}=\dfrac{21,2}{23.2+12+16.3}=0,2(mol)\\m_{HCl}=200.7,3\%=14,6(g)\\n_{HCl}=\dfrac{14,6}{36,5}=0,4(mol)\\ n_{HCl}=\dfrac{n_{HCl}}{2}\\ \to Phan\ ung\ vua\ du\\mdd_{spu}=200+200-0,2.(12+16.2)=391,2(g)\\n_{NaCl}=n_{Na_2CO_3}=0,2(mol)\\m_{NaCl}=0,4.(23+35,5)=23,4(g)\\\%m_{NaCl}=\dfrac{23,4}{391,2\%}=5,98\%$