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Chapter 17: Problem 45

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
The amount of CaCO3 that will dissolve in \(3.0 \times 10^{2} mL\) of \(0.050 M\) Ca(NO3)2 solution is approximately 1.50 grams.

Step by step solution

01

Determine the moles of Ca(NO3)2

Start by calculating the moles of Ca(NO3)2 using the formula: \[ moles = Molarity \times Volume \] where Molarity is given as \(0.050M\) and Volume is \(3.0 \times 10^{2} mL\) or \(0.3 L\). So, \[ moles = 0.050 M \times 0.3 L = 0.015 moles \]
02

Determine the moles of CaCO3

The moles of CaCO3 that can dissolve in the solution is equal to the moles of Ca(NO3)2 since one unit of Ca(NO3)2 provides one unit of Ca2+ that can combine with CO3(2-) to form a unit of CaCO3. Therefore, moles of CaCO3 will be equal to the moles of Ca(NO3)2, which is 0.015 moles.
03

Convert moles of CaCO3 to grams

Use the molar mass of CaCO3 to convert moles to grams. The molar mass of CaCO3 is \(100.09 g/mol\). Using the formula: \[ grams = moles \times molar mass \], the grams of CaCO3 that will dissolve in the solution is: \[ 0.015 moles \times 100.09 g/mol = 1.50 g \]

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Most popular questions from this chapter

The \(\mathrm{p} K_{\mathrm{a}}\) of butyric acid (HBut) is 4.7 . Calculate \(K_{\mathrm{b}}\) for the butyrate ion (But \(^{-}\) ).

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