Amount of hydrochloric acid required to consume 100g calcium carbonate completely is
(H -1U, Cl - 35.5U, C- 12, Ca – 40U) ( )
A) 7.3g B) 73g C) 0.73g D) 730g
Answers
Answered by
3
Answer:
That is for every mole of calcium carbonate reacting, there are two moles of HCl reacting. Molar mass of calcium carbonate
Answered by
17
Answer:
73g
Explanation:
CaCO3 + 2HCl = CaCl2 = H2O + CO2
Molecular weight of CaCO3 = 100g
Molecular weight of HCl = 36.5g
as there are two moles of HCl therefore, 2 x 36.5 = 73g
100g of CaCO3 consume 73g of HCl
100g of CaCO3 consume xg of HCl
x = (73 x 100)/100
x = 73g
Amount of hydrochloric acid required to consume 100g calcium carbonate completely is 73g
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