Question

use the following information to answer the next question
limewater, ( mathrm{ca}(mathrm{oh})_{2(mathrm{aq})} ) is used as a test for the presence of carbon

• dioxide gas. the reaction which takes place is:

( mathrm{ca}(mathrm{oh})_{2(mathrm{aq})}+mathrm{co}_{2(mathrm{~g})}
ightarrow mathrm{caco}_{3(mathrm{~s})}+mathrm{h}_{2} mathrm{o}_{(mathrm{l})} )

1. the amount of carbon dioxide required to precipitate

6.20 g of calcium carbonate is
16.1 mol
2.73 mol
0.0619 mol
0.141 mol

Explanation:

Step1: Calculate the molar mass of \(CaCO_3\)

The molar mass of \(CaCO_3\) is \(M = 40+12 + 3\times16=100\space g/mol\)

Step2: Find the number of moles of \(CaCO_3\)

Using the formula \(n=\frac{m}{M}\), where \(m = 6.20\space g\) and \(M = 100\space g/mol\). So \(n=\frac{6.20}{100}=0.062\space mol\)

Step3: Use the stoichiometry of the reaction

From the reaction \(Ca(OH)_{2(aq)}+CO_{2(g)}=CaCO_{3(s)}+H_2O_{(l)}\), the mole ratio of \(CO_2\) to \(CaCO_3\) is \(1:1\). So the number of moles of \(CO_2\) is equal to the number of moles of \(CaCO_3\)

Answer:

0.0619 mol (closest to 0.062 mol)