# 10 b-e 10. Use the equation below to calculate the following:\[ 3 \, \text{Ca(NO}_3\text{)}_2 \, (\text{aq}) + 2 \, \text{Na}_3\text{PO}_4 \, (\text{aq}) \rightarrow 1 \, \text{Ca}_3(\text{PO}_4)_2 \, (\text{s}) + 6 \, \text{NaNO}_3 \, (\text{aq}) \]a. Balance the reaction.b. Moles of \(\text{Na}_3\text{PO}_4\) needed to react with 1.45 L of 0.225 M \(\text{Ca(NO}_3\text{)}_2\)c. Volume of 0.250 M \(\text{Na}_3\text{PO}_4\) needed to react with 15.00 mL of 0.500 M \(\text{Ca(NO}_3\text{)}_2\)d. Grams of \(\text{Ca}_3(\text{PO}_4)_2\) that can be obtained from 125 mL of 0.500 M \(\text{Ca(NO}_3\text{)}_2\)e. Molarity of \(\text{Ca(NO}_3\text{)}_2\) when 40.00 mL react with 50.00 mL of 2.00 M \(\text{Na}_3\text{PO}_4\)

The balanced reaction is given as, 3 Ca(NO3)2 (aq) + 2 Na3PO4 (aq) →Ca3(PO4)2 (s) + 6 NaNO3 (aq)Part B: Moles of Ca(NO3)2 = Molarity x VolumeMoles of Ca(NO3)2 = 0.225 x 1.45Moles of Ca(NO3)2 = 0.326 moles3 moles of Ca(NO3)2 react = 2 moles of Na3PO40.326 moles of Ca(NO3)2 react with = 23 x 0.326 moles of Na3PO40.326 moles of Ca(NO3)2 react with =0.218 moles of Na3PO4Part C: M1V1n1 = M2V2n20.25 x V12 = 0.50 x 153V1 = 20 mlPart D: Moles of Ca(NO3)2 = Molarity x VolumeMoles of Ca(NO3)2 =0.5 x 0.125Moles of Ca(NO3)2 =0.06253 moles of Ca(NO3)2 produces = 1 mole of Ca(PO4)20.0625 moles of Ca(NO3)2 produce = 13 x 0.0625 mole of Ca3(PO4)20.0625 moles of Ca(NO3)2 produce = 0.0208 mole of Ca3(PO4)2Mass of Ca3(PO4)2 = No. of moles x Molecular MassMass of Ca3(PO4)2 =0.0208 x 310Mass of Ca3(PO4)2 =6.44 gPart E: M1V1n1 = M2V2n240 x M13 = 50 x 22M1 = 3.75 M

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Chemistry

10. Use the equation below to calculate the follow 1 Ca3(PO4)2 (s) +_ NaNO3 (aq) a. Balance the reaction. 3 Ca(NO3)2 (aq) + 2 Na3PO4 (aq) b. Moles of Na;PO4 needed to react with 1.45 L of 0.225 M Ca(NO3)2 218 mot Naz PO, c. Volume of 0.250 M Na3PO4 needed to react with 15.00 mL of 0.500 M Ca(NO3)2 2o.Ome Ne Pou d. Grams of Caz(PO4)2 that can be obtained from 125 mL of 0.500 M Ca(NO3)2 u HUa CaslPO4)n e. Molarity of Ca(NO3)2 when 40.00 mL react with 50.00 mL of 2.00 M Na3PO4

10. Use the equation below to calculate the follow 1 Ca3(PO4)2 (s) +_ NaNO3 (aq) a. Balance the reaction. 3 Ca(NO3)2 (aq) + 2 Na3PO4 (aq) b. Moles of Na;PO4 needed to react with 1.45 L of 0.225 M Ca(NO3)2 218 mot Naz PO, c. Volume of 0.250 M Na3PO4 needed to react with 15.00 mL of 0.500 M Ca(NO3)2 2o.Ome Ne Pou d. Grams of Caz(PO4)2 that can be obtained from 125 mL of 0.500 M Ca(NO3)2 u HUa CaslPO4)n e. Molarity of Ca(NO3)2 when 40.00 mL react with 50.00 mL of 2.00 M Na3PO4

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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