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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 Caâ‚‚(POâ‚„)â‚‚ give us 1 mol Pâ‚„.

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 Caâ‚‚(POâ‚„)â‚‚= 2 x (3 x (40g) + 2 x (31g) + 8 x (16 g))

2 mol Caâ‚‚(POâ‚„)â‚‚= 620 g

1 mol Pâ‚„= 4 x (31g) = 124 g

So, we can obtain 124 g of Pâ‚„ from 620 g Caâ‚‚(POâ‚„)â‚‚ (this is our recipe!)

Now, if the 75% of 1kg (=1000g) is Caâ‚‚(POâ‚„)â‚‚, we have:

1000 g x 75/100= 750 g Caâ‚‚(POâ‚„)â‚‚

Finally:

620 g Caâ‚‚(POâ‚„)â‚‚ ---------------------- 124 g Pâ‚„

750 g Caâ‚‚(POâ‚„)â‚‚ ------------------------X= (750 g x 124 g)/620 g= 150 g

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