What volume of carbon monoxide gas at STP is produced from 5.14g calcium phosphate and 6.24g silicon dioxide according , according the following equation:
2Ca3(PO4)s + 6SiO2 + 10c --> P4 + 6CaSiO3 + 10CO
a. 3.88 L
b. 8.29 x 10-2 L
c. 1.31 L
d. 1.86 L
2Ca3(PO4)s + 6SiO2 + 10c --> P4 + 6CaSiO3 + 10CO
a. 3.88 L
b. 8.29 x 10-2 L
c. 1.31 L
d. 1.86 L
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2 Ca3(PO4)2 + 6 SiO2 + 10 C → P4 + 6 CaSiO3 + 10 CO
(5.14 g Ca3(PO4)2) / (310.1782 g/mol) = 0.01657 moles Ca3(PO4)2
(6.24 g SiO2) / (60.0844 g/mol) = 0.1039 moles SiO2
0.01657 moles of Ca3(PO4)2 would react completely with (6/2) x 0.01657 = 0.04971 moles of SiO2, but there is more SiO2 than that present, so SiO2 is in excess and Ca3(PO4)2 is the limiting reactant.
(0.01657 mol Ca3(PO4)2 ) x (10/2) x (22.4 L/mol) = 1.86 L CO at STP
So answer d.
(5.14 g Ca3(PO4)2) / (310.1782 g/mol) = 0.01657 moles Ca3(PO4)2
(6.24 g SiO2) / (60.0844 g/mol) = 0.1039 moles SiO2
0.01657 moles of Ca3(PO4)2 would react completely with (6/2) x 0.01657 = 0.04971 moles of SiO2, but there is more SiO2 than that present, so SiO2 is in excess and Ca3(PO4)2 is the limiting reactant.
(0.01657 mol Ca3(PO4)2 ) x (10/2) x (22.4 L/mol) = 1.86 L CO at STP
So answer d.