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Given the following balanced reaction equation: 3 Ca(NO3)2 + 2 Li3PO4 (Molar Mass: 164.09 g/mol 115.79 g/mol If 5.58 g of calcium nitrate react with

Given the following balanced reaction equation: 3 Ca(NO3)2 + 2 Li3PO4 (Molar Mass: 164.09 g/mol 115.79 g/mol If 5.58 g of calcium nitrate react with 2.78 g of lithium phosphate, a) What is the limiting reactant in the formation of lithium nitrate? Clearly show your work. (2.5 pts) 6 LINO3 + Ca3(PO4)2 68.95 g/mol 310.18 g/mol) b) Calculate the theoretical yield (in grams) of lithium nitrate. Clearly show your work. (2.5 pts) c) Calculate the percent yield of lithium nitrate if the amount of lithium nitrate produ Clearly show your work. (2.5 pts) d) How many grams of the excess reactant were leftover? Clearly show your work. (2.5 pts)

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a To determine the limiting reactant we need to calculate the moles of each reactant and compare them Given information 558 g of CaNO32 278 g of Li3PO4 Molar mass of CaNO32 16409 gmol Molar mass of Li3PO4 11579 gmol Step 1 Calculate the moles of CaNO32 Moles of CaNO32 558 g 16409 gmol 0034 mol Step 2 Calculate the moles of Li3PO4 Moles of Li3PO4 278 g 11579 gmol 0024 mol Step 3 Compare the ... blur-text-image

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