设石灰石纯度为X,盐酸浓度为Y 溶液溶质的ω为Z
(1)有关系式得CaCO3+2HCl = CO2 + Cacl2
100 73 44 129
22*X 100*y 8.8g (22*X +100g-8.8 g)Z
解得X=10/11 约为91% Y=14.6% Z=23%
设石灰石纯度为X,盐酸浓度为Y 溶液溶质的ω为Z
(1)有关系式得CaCO3+2HCl = CO2 + Cacl2
100 73 44 129
22*X 100*y 8.8g (22*X +100g-8.8 g)Z
解得X=10/11 约为91% Y=14.6% Z=23%