Respuesta :
Phosphorus can be prepared from calcium phosphate by the following reaction:
[tex]2Ca_3(PO_4)_2(s)+6SiO_2(s)+10C(s)\rightarrow 6CaSiO_3(s)+P_4(s)+ 10CO(g)[/tex]
Phosphorite is a mineral that contains [tex]Ca_3(PO_4)_2[/tex] plus other non-phosphorus-containing compounds. What is the maximum amount of [tex]P_4[/tex] that can be produced from 2.3 kg of phosphorite if the phorphorite sample is 75% [tex]Ca_3(PO_4)_2[/tex] by mass? Assume an excess of the other reactants.
Answer: Thus the maximum amount of [tex]P_4[/tex] that can be produced is 0.345 kg
Explanation:
Given mass of phosphorite [tex]Ca_3(PO_4)_2[/tex] = 2.3 kg
As given percentage of phosphorite [tex]Ca_3(PO_4)_2[/tex] is = [tex]\frac{75}{100}\times 2.3kg=1.725kg=1725g[/tex]
[tex]moles=\frac{\text {given mass}}{\text {Molar mass}}[/tex]
[tex]{\text {moles of}Ca_3(PO_4)_2=\frac{1725g}{310g/mol}=5.56moles[/tex]
[tex]2Ca_3(PO_4)_2(s)+6SiO_2(s)+10C(s)\rightarrow 6CaSiO_3(s)+P_4(s)+ 10CO(g)[/tex]
According to stoichiometry:
2 moles of phosphorite gives = 1 mole of [tex]P_4[/tex]
Thus 5.56 moles of phosphorite give= [tex]\frac{1}{2}\times 5.56=2.78moles[/tex] of [tex]P_4[/tex]
Mass of [tex]P_4=moles\times {\text {Molar mass}}=2.78mol\times 124g/mol=345g=0.345kg[/tex]
Thus the maximum amount of [tex]P_4[/tex] that can be produced is 0.345 kg
The maximum amount of that can be produced from 2.3 kg of phosphorite in the case when sample is 75% by mass should be 0.345 kg.
Calculatoin of the maximum amount:
here moles should be
= (75% of 2.3 kg) / 310
= 1725g/310
= 5.56 moles
Now the maximum amount should be
= (50 of 556) * 124
= 345 g
= 0.345 kg
hence, The maximum amount of that can be produced from 2.3 kg of phosphorite in the case when sample is 75% by mass should be 0.345 kg.
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