Question

PLEASE ANSWER FAST ..................

CaCO₃ reacts with aqueous HCl to give CaCl₂ and CO₂. What mass of CaCO₃ is required to react completely with 25 ml of 0.70 M HCl.

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Answer 1

Balanced Equation = CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

1000 ml of 0.7 M HCl contains 0.7 * 36.5 = 25.55 g

=> 25 ml contains = 25.55 / 1000 ml * 25 g = 0.63875 g

According to equation,

2 moles of HCl reacts with 1 mole of Calcium Carbonate.

2 moles = 2 * 36.5 = 73 g of HCl

1 mole of CaCO₃ = 40 + 12 + 48 = 100 g

73 g of HCl => 100 g then, 0.63875 g of HCl => ? g of CaCO₃

Cross multiplying we get

73 * x = 0.63875 * 100

73 x = 63.875

x = 63.875 / 73 = 0.875 g of CaCO₃

Hence 0.875 g of Calcium Carbonate is needed.