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Step 1: Calculate the molar mass of H₃PO₄ and LiOH.
The molar masses of H, P, O, and Li are 1.008 g, 30.974 g, 15.999 g, and 6.94 g, respectively.
• For H₃PO₄
[tex]\begin{aligned} MM_{\text{H}_3\text{PO}_4} & = \text{3(1.008 g) + 30.974 g + 4(15.999 g)} \\ & = \text{97.994 g} \end{aligned}[/tex]
• For LiOH
[tex]\begin{aligned} MM_{\text{LiOH}} & = \text{6.94 g + 15.999 g + 1.008 g} \\ & = \text{23.95 g} \end{aligned}[/tex]
Step 2: Calculate the mass of H₃PO₄ reacted with LiOH.
Based on the balanced chemical equation, 3 moles of LiOH is stoichiometrically equivalent to 1 mole of H₃PO₄.
[tex]\begin{aligned} \text{mass of} \: \text{H}_3\text{PO}_4 & = \text{7.17 g LiOH} \times \frac{\text{1 mol LiOH}}{\text{23.95 g LiOH}} \times \frac{\text{1 mol} \: \text{H}_3\text{PO}_4}{\text{3 mol LiOH}} \times \frac{\text{97.994 g} \: \text{H}_3\text{PO}_4}{\text{1 mol} \: \text{H}_3\text{PO}_4} \\ & = \boxed{\text{9.78 g}} \end{aligned}[/tex]
Hence, 9.78 g of H₃PO₄ would actually react with 7.17 g of LiOH.