What volume of gas is evolved at S.t.p if 2g of calcium trioxocarbonate (IV) is added to a solution of hydrochloric acid?
[Ca = 40, C = 12, O = 16, Cl = 35.5, H = 1 Molar volume of a gas at s.t.p = 22.4 dm\(^3\)
112 cm3
224 cm3
448 cm3
2240 cm3
Explanation
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CaCO3(s) + 2HCl(aq) → CaCl2 + H2O(l) + CO2
The relative molecular mass of CaCO3 = 40 + 12 + 16 * 3 = 52 +48 = 100 ; Relative molecular mass of CO2 = 12 + 10 * 2 = 44
From the above equation, 100g of CaCO3reacts with 44g of CO2 at s.t.p. i.e 22.4 dm3
mole of CaCO3 = mass/molar mass
mole = 2/100=0.02mol of CaCO3
likewise the volume of gas at s.t.p is mole =Volume/22.4
where mole=0.02mol
volume=mole•volume at s.t.p
volume of the gas =0.02•22.4
volume =0.448dm3.
due to the options which are in Cm^3 we change to Cm^3 by multiplying by 1000.
0.448•1000=448cm^3.
OPTION C (448Cm^3).
_By Medic Xylene_

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CaCO3 + 2HCL >>>>>>CaCl2 + CO² + H2O
100g of CaCO3 >>>2*35.5g HCl
2g of CaCO3>>>xg of HCl
then Cross multiply
100*x=2*2*35.5
100x=142
x=1.42 g of HCL
since no of moles = m/M
then n=1.42/71
n=0.02mole
now let's find the volume
V=n*GMV(gas molar volume 22.4dm³)
V= 0.02 *22.4
V=0.448dm³ or 448cm³



