J.R. S. answered • 01/19/22
Ph.D. in Biochemistry--University Professor--Chemistry Tutor
Previously asked BY YOU, and previously answered!
See the procedure below and simply change the 12.354 g Ca(NO3)2 to 11.917 g and change the 16.076 g Rb3PO4 to 16.136 g. We aren't here to do all of your homework for you.
First, write the correctly balanced equation for the reaction taking place.
3Ca(NO3)2(aq) + 2Rb3PO4(aq) ==> Ca3(PO4)2(s) + 6RbNO3(aq) ... balanced equation
Next, we'll find which reactant is in limiting supply. An easy way to do this is to simply divide the mols of each reactant by its coefficient in the balanced equation.
For Ca(NO3)2: 12.354 g Ca(NO3)2 x 1 mol / 164.09 g = 0.07529 mols (÷3->0.0251)
For Rb3PO4: 16.076 g Rb3PO4 x 1 mol / 351.38 g = 0.04575 mols (÷2->0.0229)
Therefore, since 0.045 is less than 0.075, Rb3PO4 is the limiting reactant and will dictate how much product is formed. We will use the moles of Rb3PO4 to find the moles of Ca3(PO4)2 and then convert that to grams.
0.04575 mols Rb3PO4 x 1 mol Ca3(PO4)2 / 3 mols Rb3PO4 = 0.01525 mols Ca3(PO4)2 formed
0.01525 mols Ca3(PO4)2 x 310.18 g / mol = 4.370 g Ca3(PO4)2 formed
