Question

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

calculate the volume (in ml) of 0.100 m na₂co₃ needed to produce 1.00 g of caco₃(s)
. there is an excess of cacl₂.
molar mass of calcium carbonate = 100.09 g/mol
volume of sodium carbonate = ml

Explanation:

Step1: Calculate moles of \(CaCO_3\)

Use the formula \(n=\frac{m}{M}\), where \(m = 1.00\space g\) and \(M=100.09\space g/mol\).
\(n_{CaCO_3}=\frac{1.00\space g}{100.09\space g/mol}\approx0.01\space mol\)

Step2: Determine moles of \(Na_2CO_3\)

From the balanced equation \(Na_2CO_3(aq)+CaCl_2(aq)\to2NaCl(aq)+CaCO_3(s)\), the mole ratio of \(Na_2CO_3\) to \(CaCO_3\) is \(1:1\). So \(n_{Na_2CO_3}=n_{CaCO_3} = 0.01\space mol\)

Step3: Calculate volume of \(Na_2CO_3\)

Use the formula \(V=\frac{n}{C}\), where \(n = 0.01\space mol\) and \(C = 0.100\space M\)
\(V=\frac{0.01\space mol}{0.100\space mol/L}=0.1\space L\)
Convert liters to milliliters: \(V = 0.1\times1000\space mL = 100\space mL\)

Answer:

\(100\space mL\)