Asked by struggle
calculate the masses of each substance formed when 10g of CaCO3 reacts with excess HCl
so far I've found the ratio between the substances but from then on, I'm lost
so far I've found the ratio between the substances but from then on, I'm lost
Answers
Answered by
Olereb48
10g CaCO3 = (10/100)mole = 0.10 mole CaCO3
CaCO3 + Excess HCl =>CaCl2 + CO2 + H2O
0.10-mol CaCO3 + <HCL> => 0.10-mol CaCl2 + 0.10-mol CO2 + 0.10-mol H2O
=> 0.10(111)g CaCl2 + 0.10(44)g CO2 + 0.10(18)g H2O
=> 11.1g CaCl2 + 4.4g CO2 + 1.8g H2O
CaCO3 + Excess HCl =>CaCl2 + CO2 + H2O
0.10-mol CaCO3 + <HCL> => 0.10-mol CaCl2 + 0.10-mol CO2 + 0.10-mol H2O
=> 0.10(111)g CaCl2 + 0.10(44)g CO2 + 0.10(18)g H2O
=> 11.1g CaCl2 + 4.4g CO2 + 1.8g H2O
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