Question

please say this answer

Answer 1

Balanced chemical equation is,

CaCO3 (s) + 2HCl (aq) →  CaCl2 (aq) + H2O (l) + CO2 (g)

(40+12+3x16)U + 2(1+35.5)U → (40+2x35.5)U + (2x1+16)U + (12+2x16)U

100U + 73U → .... + ..... + 44U

As per the metric staichio equation (stoichiometric equation), 100g of CaCO3 requires 73g of HCl to liberate 44g of CO2

So, 50g (half of 100g) of CaCO3 requires 36.5 g (half of 73g) of HCl for full reaction to happen,

But only 7.3g of HCl is available to react with calcium carbonate. So on full consumption of HCl, some calcium carbonate is left over.

That is to say, HCl is available in limited quantity and it limits the amount of product formed.

Hence HCl is the limiting reagent in the given problem. (See limiting reagent)

73g of HCl gives 44g of CO2

7.3g of HCl gives (7.3/73) x 44 = 4.4g of CO2

44g of CO2 (molar mass of CO2) occupies 22.4 litre at STP

4.4g of CO2 occupies 4.4/44x22.4 = 2.24 litre at STP.

Number of molecules calculations

44g of CO2 contains 6.023x1023 mole of CO2

4.4g of CO2 contains 4.4/44x6.023x1023 = 6.023x1022 mol

Answer 2

answer is 6.022×10^22 molecules