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Phosphorous is prepared according to the following equation: Ca3(PO4)2 + 3SiO2 + 5C = 3CaSiO3 +5CO + 2P
What mass of phosphate rock (Ca3(PO4)2) is necessary to produce 9700. kg of phosphorous?

Respuesta :

unknown767

Answer:

48546.7kg or 48546745g

Explanation:

so from this u can work out the amount of moles in phosphorous by doing mass / mr you have to convert the mass in kg to g so you times it by 1000. then divide it by 31 which is the mr of phosphorous. then u can use the molar ratio which is 2:1 . Then use the equation mass= moles*mr

so 97000*1000/31=312903moles

2:1

so 312903/2=156451 moles of ca3(PO4)2

so mass= moles*mr

156451*310.3=48494g

hope this helps u to understand(*°▽°*)

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