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 cm2
- 224 cm2
-
448 cm2
- 2240 cm2
The correct answer is: C
Explanation
CaCO\(_3\)(s) + 2HCl(aq) → CaCl\(_2\) + H\(_2\)O(l) + CO\(_2\)
The relative molecular mass of CaCO\(_3\) = 40 + 12 + (16 x 3) = 52 + 48 = 100 ;
Relative molecular mass of CO\(_2\)= 12 + (10 x 2) = 44
From the above equation, 100g of CaCO\(_3\) reacts with 44g of CO\(_2\) at s.t.p. i.e 22.4 dm\(^3\)
2g of CaCO\(_3\) will react with (2 x 22.4)/100 of CO\(_2\) = 0.448 dm\(^3\)
0.448 dm\(^3\) = 0.448 x 1000 = 448cm\(^3\)
Alternatively,
CaCO\(_3\)(s) + 2HCl(aq) → CaCl\(_2\) + H\(_2\)O(l) + CO\(_2\)
1 : 2 1
The focus is on CaCO\(_3\) and CO\(_2\) and they are in ratio 1 : 1, hence, we can calculate the number of moles of CaCO\(_3\) and substitute the value for the number of moles of CO\(_2\), since their number of moles will also be the same.
Molar mass of CaCO\(_3\) = Ca + C + O\(_3\) = 40 + 12 + (16 x 3) = 100g/mol
Number of moles of CaCO\(_3\) = \(\frac{mass}{molar mass}\) = \(\frac{2}{100}\) = 0.02 mole
Number of moles = \(\frac{Volume}{molar volume}\)
0.02 = \(\frac{volume}{22.4}\)
Volume = 0.02 x 22.4 = 0.02 x 22400 = 448 cm\(^3\)