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QUESTION

# Balance the following reaction: CaCO3 + FePO4 --&gt; Ca3(PO4)2 + Fe2(CO3)3. Which reactant is limiting, assuming we start with 100 grams of calcium carbonate and 45 grams of iron (III) phosphate. What is the mass of each product that can be formed? What m...

3CaCO_3+ 2FePO_4 --> Ca_3(PO_4)2 + Fe_2(CO_3)3 (a)

As per the above equation (a) Three moles of CaCO_3 , consumes two moles of FePO_4.

In terms of mass one mole of CaCO_3 , has mass 100 g/mol. and one mole of FePO_4 has mass 150.8 g/mol.

so let us set up the ratio;

("3 mole" CaCO_3)/("2 mole" FePO_4) = (300g)/(301.6g) (b)

X g of CaCO_3 will consume 45 g of FePO_4 (c)

equating two equation (b) and (c)

(300g)/(301.6g) = "Xg"/"45g"

300 x 45 = 301.6 X

301.6 X = 13500

X = 13500/301.6 = 44.7 g => 45 g

So, 45 g of CaCO_3 will react with 45 g of FePO_4 . The amount of FePO_4 added is 45 g. The amount of CaCO_3 remains unused is 100-45= 55 g. Iron (III) phosphate is a .

2 moles of FePO_4 produces 1 mole of Ca_3(PO_4)2

301.6 g of FePO_4 produces 310.17 g Ca_3(PO_4)2 (d)

45 FePO_4 produces X g Ca_3(PO_4)2 (e)

"301.6g"/"310.17g" = "45g"/"Xg"

301.6 ( X ) = 45 x 310.17

301.6 (X) = 13957.65

X = 13957.65/301.6 = 46.27 g