Sr(OH)_{2} (aq) + 2 HNO_{3} (aq) → Sr(NO_{3})_{2} (aq) + 2 H_{2}O (l)

Step 1

$\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{Sr}{\left(\mathrm{OH}\right)}_{\mathbf{2}}}\mathbf{}\mathbf{\times}\mathbf{}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{Sr}{\left(\mathrm{OH}\right)}_{\mathbf{2}}}{\mathbf{121}\mathbf{.}\mathbf{63}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{Sr}{\left(\mathrm{OH}\right)}_{\mathbf{2}}}\mathbf{}}$** = 0.1233 mol Sr(OH) _{2}**

**$\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{}\overline{)\mathbf{mL}}\mathbf{}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\overline{)\mathbf{L}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mL}}}\mathbf{\times}\frac{\mathbf{0}\mathbf{.}\mathbf{210}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{HNO}}_{\mathbf{3}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{L}}}$ = 0.0126 mol HNO _{3}**

A solution is made by mixing 15.0 g of Sr(OH)_{2} and 60.0 mL of 0.210 M HNO_{3}.

Calculate the concentration of Sr^{2+} ion remaining in solution.

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