Question

what volume of 1 molarity HCL solution is required to dissolve 0.25 gram of csu 3 and how many ml of Co2 at STP will be evolved during the reaction​

Answer 1

Answer:

The molecular weight of Calcium Carbonate is 100.

(CaCO3 : 40 + 12 + 16*3 = 100)

That means 1 mole of Calcium Carbonate weighs 100g.

Hence 1g will contain 0.01 moles of Calcium Carbonate.

CaCO3 + Acid - - - > Salt + CO2 + Water

Hence on complete neutralisation, 0.01 moles of carbon dioxide will be formed.

1 mole of a gas occupies 22.4L at STP.

Therefore, 0.224L or 224ml of CO2 will be formed.