Question

3:you place 2.56 g of caco3(fw = 100.09 g/mol) in a beaker containing 250. Ml of 0.125 m hcl. What the reactions has ceased, what mass of cacl2(fw = 110.98 g/mol) can be produced? The unbalancedequation i

Answer 1

The balanced chemical equation between calcium carbonate and hydrochloric acid:

$CaCO_{3}(s)+2HCl(aq)-->CaCl_{2}(aq)+H_{2}O(l)+CO_{2}(g)$

Moles of $CaCO_{3}$=$2.56g*\frac{1mol}{100.09g}=0.0256mol CaCO_{3}$

Moles of HCl=$250mL*\frac{1L}{1000mL}*\frac{0.125mol}{L} =0.03125mol HCl$

The mole ratio of $\frac{CaCO_{3} }{HCl} =\frac{1}{2}$

But we have $0.0256 mol CaCO_{3}$ per 0.03125 mol HCl

Hence, HCl is the limiting reactant. The amount of product formed would depend on moles of HCl.

Mass of $CaCl_{2}$=0.03125mol HCl *$\frac{1mol CaCl_{2} }{2mol HCl} *\frac{110.98 g}{1 mol}=1.73 g CaCl_{2}$