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(B) $1$. Balanced chemical equation: $Ca_3(PO_4)_2 + 3SiO_2 + 5C + 5O_2 + 3H_2O 	o 3CaSiO_3 + 5CO_2 + 2H_3PO_4$$2$. Molar mass of $Ca_3(PO_4)_2 = 3(4

Vedclass · Accepted Answer

$0.63$

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(B) $1$. Balanced chemical equation: $Ca_3(PO_4)_2 + 3SiO_2 + 5C + 5O_2 + 3H_2O 	o 3CaSiO_3 + 5CO_2 + 2H_3PO_4$$2$. Molar mass of $Ca_3(PO_4)_2 = 3(40) + 2(31 + 64) = 120 + 190 = 310 \ g/mol$.$3$. Molar mass of $SiO_2 = 28 + 32 = 60 \ g/mol$.$4$. Moles of $Ca_3(PO_4)_2 = 1000 \ g / 310 \ g/mol \approx 3.226 \ mol$.$5$. Moles of $SiO_2 = 1000 \ g / 60 \ g/mol \approx 16.667 \ mol$.$6$. According to the stoichiometry,$1 \ mol$ of $Ca_3(PO_4)_2$ requires $3 \ mol$ of $SiO_2$. Thus,$3.226 \ mol$ of $Ca_3(PO_4)_2$ requires $3.226 	imes 3 = 9.678 \ mol$ of $SiO_2$.$7$. Since we have $16.667 \ mol$ of $SiO_2$ (which is $> 9.678 \ mol$),$Ca_3(PO_4)_2$ is the limiting reagent.$8$. $1 \ mol$ of $Ca_3(PO_4)_2$ produces $2 \ mol$ of $H_3PO_4$. Therefore,$3.226 \ mol$ of $Ca_3(PO_4)_2$ produces $3.226 	imes 2 = 6.452 \ mol$ of $H_3PO_4$.$9$. Molar mass of $H_3PO_4 = 3(1) + 31 + 4(16) = 98 \ g/mol$.$10$. Mass of $H_3PO_4 = 6.452 \ mol 	imes 98 \ g/mol \approx 632.3 \ g \approx 0.63 \ kg$.

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