What volume of carbon monoxide gas at STP is produced from 6.14 g calcium phosphate and...?
2Ca3 (PO4)2 + 6SiO2 + 10C —> P4 + 6CaSiO3 + 10CO (already balanced)
Answer:
Firstly convert the masses in forms of number of moles.
Molecular weight of Calcium phosphate = 310.18 grams/mole
Number of moles of Calcium phosphate involved = 6.14/310.18 = 0.0198 moles
Molecular weight of SiO2 = 60.1 grams/mole
Number of moles of SiO2 involved = 8.24/60.1 = 0.1371 moles
They react in the ratio 1:3 by moles. So, the limiting reagent is calcium phosphate. All of it will be used up.
According to the equation, 2 moles of calcium phosphate produce 10 moles of CO. So, we have 0.099 moles of CO formed from the 0.0198 moles of Calcium phosphate reacted.
Each mole of any gas occupies 22.4 L at STP. So, the volume occupied by CO = 2.2176 L at STP.
if 620g(molecular weight)of 2Ca3(PO4)2 gives 10*22.4L of CO(22.4L is molar volume and there are 10 moles)
then 6.14g of 2Ca3(PO4)2 gives xL of CO.
x=2.7729L of Carbon monoxide.
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