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Ca3po42 sio2 c casio3 co p balanced?
Ca3(PO4)2 + 3 SiO2 + 5 C = 3 CaSiO3 + 5 CO + 2 P
How many atoms does Ca3PO42?
This appears to be calcium triphosphate, Ca3(PO4)2. In any formula such as this when an element appears without a subscript this means that just one atom of the element is to be understood. Thus we could rewrite this formula as Ca3(P1O4)2. 3 Ca 2 P 8 O -- 13 atoms altogether
What is the maximum amount of P4 that can be produced from 1.0kg of phosphorite if the phorphorite sample is 75 Ca3PO42 by mass?
2Ca3(PO4)2 + 6SiO2 + 10 C -> 6CaSiO3 + P4 + 10 CO There is 1 kg of solution. 75% of the solution is Ca3(PO4)2. Change 1kg to grams which is 1000g. 75% of 1000 is 750 Therefore there is 750g of the Ca3(PO4)2. Find the mole of Ca3(PO4)2 which is the mass divided by the molar mass. 750g/310.4gmol-1 which equals n =2.4mol ...1/2 is the ratio so mol of P4 is 1.2 mol. to find the mass of P4 you do M * n = 1.2mol * 148.65 therefore the max amount of P4 that can be produced is 150g
What does ca3po42 stand for?
It stands for Calcium Phosphate.
Ca3po42 sio2 c casio3 co p balanced?
Ca3(PO4)2 + 3 SiO2 + 5 C = 3 CaSiO3 + 5 CO + 2 P
How many atoms does Ca3PO42?
This appears to be calcium triphosphate, Ca3(PO4)2. In any formula such as this when an element appears without a subscript this means that just one atom of the element is to be understood. Thus we could rewrite this formula as Ca3(P1O4)2. 3 Ca 2 P 8 O -- 13 atoms altogether
What is the gram formula mass for Ca3PO42?
Ca = 40 x 3 = 120 grams P = 31 x 2 = 62 grams O = 16 x 8 = 128 grams Add them together and you have 310 grams. This is the mass in grams of your formula. It is expressed in grams/mole because 310 is the weight of one mole of Ca3(PO4)2.
What is the maximum amount of P4 that can be produced from 1.0kg of phosphorite if the phorphorite sample is 75 Ca3PO42 by mass?
2Ca3(PO4)2 + 6SiO2 + 10 C -> 6CaSiO3 + P4 + 10 CO There is 1 kg of solution. 75% of the solution is Ca3(PO4)2. Change 1kg to grams which is 1000g. 75% of 1000 is 750 Therefore there is 750g of the Ca3(PO4)2. Find the mole of Ca3(PO4)2 which is the mass divided by the molar mass. 750g/310.4gmol-1 which equals n =2.4mol ...1/2 is the ratio so mol of P4 is 1.2 mol. to find the mass of P4 you do M * n = 1.2mol * 148.65 therefore the max amount of P4 that can be produced is 150g
