Question

how many ml of 0.499 m hbr are needed to dissolve 9.38 g of caco₃?
2hbr(aq) + caco₃(s) → cabr₂(aq) + h₂o(l) + co₂(g)
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ml

Explanation:

Step1: Calculate the moles of \(CaCO_3\)

The molar mass of \(CaCO_3\) is \(M = 40.08+12.01 + 3\times16.00=100.09\space g/mol\).
The moles of \(CaCO_3\), \(n=\frac{m}{M}\), where \(m = 9.38\space g\). So \(n=\frac{9.38\space g}{100.09\space g/mol}\approx0.0937\space mol\).

Step2: Determine the moles of \(HBr\) from the stoichiometry

From the balanced equation \(2HBr(aq)+CaCO_3(s)\to CaBr_2(aq)+H_2O(l)+CO_2(g)\), the mole ratio of \(HBr\) to \(CaCO_3\) is \(2:1\).
So the moles of \(HBr\), \(n_{HBr}=2\times n_{CaCO_3}\). Since \(n_{CaCO_3} = 0.0937\space mol\), \(n_{HBr}=2\times0.0937\space mol = 0.1874\space mol\).

Step3: Calculate the volume of \(HBr\) solution

The formula for molarity \(M=\frac{n}{V}\) (where \(M = 0.499\space M\) and \(n = 0.1874\space mol\)). Rearranging for \(V\), \(V=\frac{n}{M}\).
\(V=\frac{0.1874\space mol}{0.499\space mol/L}\approx0.376\space L\).
Since \(1\space L=1000\space mL\), \(V = 0.376\times1000\space mL = 376\space mL\).

Answer:

\(376\space mL\)