Consider the following unbalanced equation: Ca3(PO4)2(s)+3H2SO4(aq)⇒3CaSO4(s)+2H3PO4(aq). What masses of calcium sulfate and phosphoric acid can be produced from the reaction of 1.0 kg of calcium phosphate with 1.0 kg of concentrated sulfuric acid (98%H2SO4 by mass)?

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Cassandra Ramirez · Accepted Answer

Balance the equation:  Ca3(PO4)2(s)+3H2SO4(aq)⇒3CaSO4(s)+2H3PO4(aq)  Solve for the masses of the reactants:  H2SO4  1.00kgH2SO4×0.98×1mol H2SO498.09×10−3kgH2SO4=10.molH2SO4  Ca3(PO4)2  1.00kgCa3(PO4)2×1mol Ca3(PO4)2310.18×10−3kgCa3(PO4)2  =3.2molCa3(PO4)2×3molH2SO41molCa3(PO4)2=9.6molH2SO4  Since 9.6mol&amp;lt;10mol,  Ca3(PO4)2  is the limiting reactant.  Step 3  Finally, solve for the masses of the products:  3.2molCa3(PO4)2×3molCaSO41molCa3(PO4)2×135.15gCaSO41molCaSO4  =1.3×103gCaSO4

zurilomk4 · Accepted Answer

First, balance the reaction. Ca3(PO4)2(s)+3H2SO4(aq)⇒3CaSO4(s)+2H3PO4(aq) The molar mass of calcium phosphate is 40.08×3+(30.97+16×4)×2=310.18gmol Therefore 1.0kg of calcium phosphate =1000g, and dividing by 310.18g/mol gives 3.22 moles. The molar mass of the sulfuric acid is 1.01×2+32.07+16×4=98.09gmol The 1.0kg mixture is 98% sulfuric acid, which is 0.98kg or 980g. Dividing by the 98.09g/mol gives 9.99 moles. From the balanced reaction, 1 mol calcium phosphate reacts with 3 mol sulfuric acid. Therefore, 3.22 mol calcium phosphate will react with 3(3.22)=9.66mol sulfuric acid. There are 9.99 mol sulfuric acid, so it is the excess reactant. We use the limiting reactant of calcium phosphate for the calculations. From the reaction, 1 mol calcium phosphate forms 3 mol calcium sulfate and 2 mol phosphoric acid. (3.22molCa3(PO4)2)(3)=9.66molCaSO4 (3.22molCa3(PO4)2)(2)=6.44molH3PO4

Consider the following unbalanced equation: Ca_{3} (PO_{4})_{2}(s)+3H_{2}SO_{4}(aq) \Rightarrow 3CaSO_{4}(s)+2H_{3}PO_{4}(aq). What

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