(2)碳酸钙与稀盐酸反应的化学方程式为:CaCO3+2HCl=CaCl2+H2O+CO2↑;
(3)设生成氯化钙质量为x,
CaCO3+2HCl=CaCl2+H2O+CO2↑
111 44
x 15g+105.4g-111g-5g
∴质量
111 |
44 |
x |
15g+105.4g−5g−111g |
解之得:x=11g;
(4)
11g |
111g |
故答案为:(1)碳酸钙;
(2)CaCO3+2HCl=CaCl2+H2O+CO2↑;
(3)
111 |
44 |
x |
15g+105.4g−5g−111g |
111 |
44 |
x |
15g+105.4g−5g−111g |
11g |
111g |
111 |
44 |
x |
15g+105.4g−5g−111g |