Correct Answer: Option 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\(^3\)

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