设CaCO3质量为x,HCl质量为y,CaCl2质量为z
CaCO3+2HCl=CaCl2+CO2↑+H20
100 73 111 44
x y z 4.4g
100:44= x:4.4g 73 :44=y :4.4g 111:44= z :4.4g
x=10g,y=7.3g,z=11.1g
(1)石灰石中CaCO3的质量分数为:10g/12.5*100%=80%
(2)稀盐酸中溶质的质量分数为:7.3g/36.5g*100%=20%
(3)所得CaCl2溶液中溶质的质量分数为:11.1g/(10+36.5g-4.4g)*100%=26.4%
答略.