CaCl, CO₂ and H₂O are produced by the reaction of dilute HCI with CaCO₃ . How many grams of HCI will be required to prepare 55.5g of CaCl, 22.2 g of CO₂ and 9.0 g of H₂O from 50.0g of CaCO₃? What is the number of mole of HCI required? [H = 1, Cl = 35.5]
Answers
Answer:
CaCO3 + 2HCl -----> CaCl2 + H2O + CO2.
2.5g? g
Mass of CaCO3= 40+12+3(16)=52+ 48= 100.
Molar mass of CaCO3=100g.
Mass of CaCl2= 40 + 2(35.5)= 40+ 71= 111.
Molar mass of CaCl2=111g.
Mass of CaCO3 (g) Mass of CaCl2 (g)
For 100g of CaCO
3
, 111g of CaCl_2$$ is formed.
let for 2.5g of CaCO
3
, x g of CaCl
2
is formed.
Thus, by cross multiplication,
x=111×2.5/100=2.775g=2.78g.
Explanation:
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Answer :-
36.7 grams of HCl is required. Also, the number of moles of HCl is nearly 1 .
Explanation :-
Let the required mass of HCl be x grams. The skeletal chemical equation for the given reaction is :-
CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O
In this reaction CaCO₃ and HCl are reactants whereas CaCl₂ , CO₂ and H₂O are products. Now, according to "Law of Conservation of mass", we know that :-
Mass of reactants = Mass of products
⇒ 50 + x = 55.5 + 22.2 + 9
⇒ 50 + x = 86.7
⇒ x = 86.7 - 50
⇒ x = 36.7 grams
________________________________
Molar mass of HCl is (35.5 + 1) = 36.5 g/mol.
⇒ Moles = Given mass/Molar mass
⇒ Moles = 36.7/36.5
⇒ Moles ≈ 1