Question

1. na₂co₃(aq) + cacl₂(aq) → 2nacl(aq) + caco₃(s)

calculate the volume (in ml) of 0.200 m cacl₂ needed to produce 2.00 g of caco₃(s)
. there is an excess of na₂co₃.
molar mass of calcium carbonate = 100.09 g/mol
volume of calcium chloride = ml

Explanation:

Step1: Calculate moles of \(CaCO_3\)

Use the formula \(n=\frac{m}{M}\), where \(m = 2.00\ g\) and \(M=100.09\ g/mol\).
\(n_{CaCO_3}=\frac{2.00\ g}{100.09\ g/mol}\approx0.01998\ mol\)

Step2: Relate moles of \(CaCl_2\) to \(CaCO_3\)

From the balanced equation \(Na_2CO_3(aq)+CaCl_2(aq)\to2NaCl(aq)+CaCO_3(s)\), the mole ratio \(n_{CaCl_2}:n_{CaCO_3}=1:1\). So \(n_{CaCl_2}=n_{CaCO_3}\approx0.01998\ mol\)

Step3: Calculate volume of \(CaCl_2\) solution

Use the formula \(V=\frac{n}{C}\), where \(n = 0.01998\ mol\) and \(C = 0.200\ M\)
\(V=\frac{0.01998\ mol}{0.200\ mol/L}=0.0999\ L\)

Convert liters to milliliters: \(V = 0.0999\ L\times1000\ mL/L = 99.9\ mL\)

Answer:

\(99.9\)