How many grams of \(\mathrm{CaCO}_{3}\) will dissolve in \(3.0 \times\) \(10^{2} \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} ?\)

Short Answer

Expert verified
Around \(1.50\) g of CaCO3 will dissolve in \(3.0 \times 10^{2}\) mL of \(0.050\) M Ca(NO3)2.

Step by step solution

01

Calculate the number of moles of Ca(NO3)2

The number of moles of solution is calculated by multiplying the volume of the solution by its molarity. In this case, we have \(3.0 \times 10^{2}\) mL of \(0.050\) M Ca(NO3)2. We first convert the volume to liters (as one liter equals to \(1000\) mL) and then use the formula: number of moles = molarity x volume in liters. So, the calculation becomes: number of moles = \(0.050\) mol/L x \(3.0 \times 10^{2}/1000\) L = \(0.015\) mol.
02

Apply stoichiometric ratios

The stoichiometric ratio between \(Ca(NO_3)_2\) and \(CaCO_3\) can be determined from the chemical equation representing their reaction: Ca(NO3)2 + CO2 + H2O -> CaCO3 + 2HNO3. From this balanced chemical equation, we can see that one mole of Ca(NO3)2 will produce one mole of CaCO3. As we found \(0.015\) mol of Ca(NO3)2 in the solution, we can say that it will produce \(0.015\) mol of CaCO3.
03

Calculate the mass of CaCO3

Finally, we can find the mass of CaCO3 that will dissolve in the solution by multiplying the number of moles by the molar mass of CaCO3. The molar mass of CaCO3 is approximately \(100.09\) g/mol. So, the amount of CaCO3 that will dissolve is \(0.015\) mol x \(100.09\) g/mol = \(1.50\) g.

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