решение:
CaCO3+2HCl=>CaCl2+CO2+H2O
xг(CaCO3)----------3,38л(CO2)
100г(CaCO3)--------22,4л(CO2)
х=(3,38*100)/22,4=15,09л CaCO3
28,8г-------100%.
15,09л--------х
х=(15,09*100)/28,8=52,4%
Ответ:52,4%
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решение:
CaCO3+2HCl=>CaCl2+CO2+H2O
xг(CaCO3)----------3,38л(CO2)
100г(CaCO3)--------22,4л(CO2)
х=(3,38*100)/22,4=15,09л CaCO3
28,8г-------100%.
15,09л--------х
х=(15,09*100)/28,8=52,4%
Ответ:52,4%