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Chapter 3: Problem 124

Consider the following unbalanced equation: $$ \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{CaSO}_{4}(s)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) $$ What masses of calcium sulfate and phosphoric acid can be produced from the reaction of 1.0 \(\mathrm{kg}\) calcium phosphate with 1.0 \(\mathrm{kg}\) concentrated sulfuric acid $\left(98 \% \mathrm{H}_{2} \mathrm{SO}_{4} \text { by mass)? }\right.$

Short Answer

Expert verified
The mass of calcium sulfate produced is \(1315.83~g\), and the mass of phosphoric acid produced is \(631.54~g\) when reacting 1.0 kg of calcium phosphate with 1.0 kg of concentrated sulfuric acid (98% by mass).

Step by step solution

01

Balance the equation.

First, we need to balance the given chemical equation to maintain the conservation of mass: Ca3(PO4)2(s) + H2SO4(aq) → CaSO4(s) + H3PO4(aq) Balanced equation: 2Ca3(PO4)2(s) + 6H2SO4(aq) → 6CaSO4(s) + 4H3PO4(aq)
02

Calculate the moles of reactants.

Next, we will determine the amounts of reactants available in moles using their respective molar masses: Molar mass of Ca3(PO4)2 = 310.18 g/mol Molar mass of H2SO4 = 98.08 g/mol Given mass of Ca3(PO4)2 = 1.0 kg = 1000 g Given mass of H2SO4 = 1.0 kg × 98 % = 980 g Moles of Ca3(PO4)2 = mass / molar mass Moles of Ca3(PO4)2 = 1000 g / 310.18 g/mol = 3.22 mol Moles of H2SO4 = mass / molar mass Moles of H2SO4 = 980 g / 98.08 g/mol = 9.99 mol
03

Identify the limiting reactant.

We can now identify the limiting reactant by comparing the ratios of moles of reactants to the stoichiometry of the balanced equation: Mole ratio of Ca3(PO4)2 to H2SO4 in the balanced equation = 2 : 6 Divide the moles of each reactant by their stoichiometric coefficient: Ca3(PO4)2 = 3.22 mol / 2 = 1.61 H2SO4 = 9.99 mol / 6 = 1.67 Since the value for Ca3(PO4)2 is lower, it is the limiting reactant.
04

Calculate the mass of products formed.

Now, using the limiting reactant (Ca3(PO4)2), we can calculate the mass of the products formed (CaSO4 and H3PO4). Molar mass of CaSO4 = 136.14 g/mol Molar mass of H3PO4 = 97.99 g/mol From the balanced equation, we see that every 2 moles of Ca3(PO4)2 will produce 6 moles of CaSO4. Moles of CaSO4 produced = (3.22 mol Ca3(PO4)2 / 2) × 6 = 9.66 mol Mass of CaSO4 produced = moles × molar mass Mass of CaSO4 = 9.66 mol × 136.14 g/mol = 1315.83 g Similarly, every 2 moles of Ca3(PO4)2 will produce 4 moles of H3PO4. Moles of H3PO4 produced = (3.22 mol Ca3(PO4)2 / 2) × 4 = 6.44 mol Mass of H3PO4 produced = moles × molar mass Mass of H3PO4 = 6.44 mol × 97.99 g/mol = 631.54 g
05

Present the final answer.

The mass of calcium sulfate produced is 1315.83 g, and the mass of phosphoric acid produced is 631.54 g when reacting 1.0 kg of calcium phosphate with 1.0 kg of concentrated sulfuric acid (98% by mass).

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