Question #e484d

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Question #e484d

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Answer 1 · 2014-03-21T09:34:52

For this question we have to start with balanced chemical equation.

$3 C a C O_{3}$ $3CaCO_3$ + $2 F e P O_{4}$ $2FePO_4$ --> $C a_{3} (P O_{4}) 2$ $Ca_3(PO_4)2$ + $F e_{2} (C O_{3}) 3$ $Fe_2(CO_3)3$ (a)

As per the above equation (a) Three moles of $C a C O_{3}$ $CaCO_3$ , consumes two moles of $F e P O_{4}$ $FePO_4$ .

In terms of mass one mole of $C a C O_{3}$ $CaCO_3$ , has mass 100 g/mol.
and one mole of $F e P O_{4}$ $FePO_4$ has mass 150.8 g/mol.

so let us set up the ratio;

$\frac{3 mole C a C O_{3}}{2 mole F e P O_{4}}$ $("3 mole" CaCO_3)/("2 mole" FePO_4)$ = $\frac{300 g}{301.6 g}$ $(300g)/(301.6g)$ (b)

X g of $C a C O_{3}$ $CaCO_3$ will consume 45 g of $F e P O_{4}$ $FePO_4$ (c)

equating two equation (b) and (c)

$\frac{300 g}{301.6 g}$ $(300g)/(301.6g)$ = $\frac{Xg}{45g}$ $"Xg"/"45g"$

300 x 45 = 301.6 X

301.6 X = 13500

X = $\frac{13500}{301.6}$ $13500/301.6$ = 44.7 g => 45 g

So, 45 g of $C a C O_{3}$ $CaCO_3$ will react with 45 g of $F e P O_{4}$ $FePO_4$ . The amount of $F e P O_{4}$ $FePO_4$ added is 45 g. The amount of $C a C O_{3}$ $CaCO_3$ remains unused is 100-45= 55 g. Iron (III) phosphate is a limiting reagent.

2 moles of $F e P O_{4}$ $FePO_4$ produces 1 mole of $C a_{3} (P O_{4}) 2$ $Ca_3(PO_4)2$

301.6 g of $F e P O_{4}$ $FePO_4$ produces 310.17 g $C a_{3} (P O_{4}) 2$ $Ca_3(PO_4)2$ (d)

45 $F e P O_{4}$ $FePO_4$ produces X g $C a_{3} (P O_{4}) 2$ $Ca_3(PO_4)2$ (e)

$\frac{301.6g}{310.17g}$ $"301.6g"/"310.17g"$ = $\frac{45g}{Xg}$ $"45g"/"Xg"$

301.6 ( X ) = 45 x 310.17

301.6 (X) = 13957.65

X = $\frac{13957.65}{301.6}$ $13957.65/301.6$ = 46.27 g

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