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Phosphorus can be prepared from calcium phosphate by the following reaction:
2Ca₃(PO₄)₂+6SiO₂+10C --> 6CaSiO₃+P₄+10CO
Phosphorite is a mineral that contains Ca₃(PO₄)₂ plus other non-phosphorus-containing compounds. What is the maximum amount of P₄ that can be produced from 1.0 kg of phosphorite if the phorphorite sample is 75% Ca₃(PO₄)₂ by mass? Assume an excess of the other reactants.

Respuesta :

jbferrado

Answer:

150 g (0.15 kg)

Explanation:

From the equation, 2 mol Ca₂(PO₄)₂ give us 1 mol P₄.

To calculate the weight, we need the molecular weights (Mw) of the elements:

Mw Ca= 40 g

Mw P= 31 g

Mw O= 16 g

2 mol Ca₂(PO₄)₂= 2 x (3 x (40g) + 2 x (31g) + 8 x (16 g))

2 mol Ca₂(PO₄)₂= 620 g

1 mol P₄= 4 x (31g) = 124 g

So, we can obtain 124 g of P₄ from 620 g Ca₂(PO₄)₂ (this is our recipe!)

Now, if the 75% of 1kg (=1000g) is Ca₂(PO₄)₂, we have:

1000 g x 75/100= 750 g Ca₂(PO₄)₂

Finally:

620 g Ca₂(PO₄)₂ ---------------------- 124 g P₄

750 g Ca₂(PO₄)₂ ------------------------X= (750 g x 124 g)/620 g= 150 g

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