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Chapter 4: Problem 25

When \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) and \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) are combined, the following reaction occurs: $$ 2 \mathrm{PO}_{4}{ }^{3-}(a q)+3 \mathrm{Ca}^{2+}(a q) \longrightarrow \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) $$ How many grams of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)(\mathrm{MM}=310.18 \mathrm{~g} / \mathrm{mol})\) are obtained when \(15.00 \mathrm{~mL}\) of \(0.1386 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) are mixed with \(20.00 \mathrm{~mL}\) of \(0.2118 \mathrm{M}\) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} ?\)

Short Answer

Expert verified
Answer: 0.3223 g

Step by step solution

01

Find the moles of reactants

First, we need to find the moles of both reactants using their volumes and molarities. Moles of Na3PO4 = (Volume of Na3PO4) × (Molarity of Na3PO4) Moles of Na3PO4 = (15.00 mL) × (0.1386 mol/L) Moles of Na3PO4 = 0.002079 mol Moles of Ca(NO3)2 = (Volume of Ca(NO3)2) × (Molarity of Ca(NO3)2) Moles of Ca(NO3)2 = (20.00 mL) × (0.2118 mol/L) Moles of Ca(NO3)2 = 0.004236 mol
02

Identify the limiting reactant

We need to determine which reactant is the limiting reactant by calculating the mole ratios. We are given that the reaction has a stoichiometry of: 2 PO4⁻³(aq) + 3 Ca²⁺(aq) → Ca3(PO4)2(s) Mole ratio of Na3PO4 to Ca(NO3)2 = (Moles of Na3PO4) / 2 : (Moles of Ca(NO3)2) / 3 Mole ratio = (0.002079/2) : (0.004236/3) = 0.0010395 : 0.0014120 Since the mole ratio of Na3PO4 to Ca(NO3)2 is less than the stoichiometry, Na3PO4 is the limiting reactant.
03

Calculate the moles of calcium phosphate formed

Now, we will find the moles of calcium phosphate (Ca3(PO4)2) formed using the stoichiometry. 2 moles of PO4⁻³ give 1 mole of Ca3(PO4)2, and we have 0.002079 moles of Na3PO4 (which contains the PO4⁻³ ions). Moles of Ca3(PO4)2 = (0.002079 mol) × (1 mol Ca3(PO4)2 / 2 mol Na3PO4) Moles of Ca3(PO4)2 = 0.0010395 mol
04

Calculate the mass of calcium phosphate formed

Finally, we will calculate the mass of calcium phosphate (Ca3(PO4)2) formed using its molar mass, which is given as 310.18 g/mol. Mass of Ca3(PO4)2 = (Moles of Ca3(PO4)2) × (Molar Mass of Ca3(PO4)2) Mass of Ca3(PO4)2 = (0.0010395 mol) × (310.18 g/mol) Mass of Ca3(PO4)2 = 0.3223 g Thus, when 15.00 mL of 0.1386 M Na3PO4 are mixed with 20.00 mL of 0.2118 M Ca(NO3)2, 0.3223 g of Ca3(PO4)2 are obtained.

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Most popular questions from this chapter

Limonite, an ore of iron, is brought into solution in acidic medium and titrated with \(\mathrm{KMnO}_{4}\). The unbalanced equation for the reaction is $$ \mathrm{MnO}_{4}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q)+\mathrm{Mn}^{2+}(a q) $$ It is found that a \(1.000-\mathrm{g}\) sample of the ore requires \(75.52 \mathrm{~mL}\) of \(0.0205 \mathrm{M}\) \(\mathrm{KMnO}_{4}\). What is the percent of Fe in the sample?

III Classify each of the following half-reactions as oxidation or reduction. (a) \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{O}^{2-}(a q)\) (b) \(\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)\) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)\) (d) \(\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Cl}_{2}(g)\)

What is the volume of \(1.222 \mathrm{M}\) sodium hydroxide required to react with (a) \(32.5 \mathrm{~mL}\) of \(0.569 \mathrm{M}\) sulfurous acid? (One mole of sulfurous acid reacts with two moles of hydroxide ion.) (b) \(5.00 \mathrm{~g}\) of oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) ? (One mole of oxalic acid reacts with two moles of hydroxide ion.) (c) \(15.0 \mathrm{~g}\) of concentrated acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) that is \(88 \%\) by mass pure?

What is the volume of \(0.885 M\) hydrochloric acid required to react with (a) \(25.00 \mathrm{~mL}\) of \(0.288 \mathrm{M}\) aqueous ammonia? (b) \(10.00 \mathrm{~g}\) of sodium hydroxide? (c) \(25.0 \mathrm{~mL}\) of a solution \(\left(d=0.928 \mathrm{~g} / \mathrm{cm}^{3}\right)\) containing \(10.0 \%\) by mass of methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right) ?\)

Write the formulas of the following compounds and decide which are soluble in water. (a) sodium sulfate (b) iron(III) nitrate (c) silver chloride (d) chromium(III) hydroxide
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