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

Question

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?

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Sapo 4 years 2021-07-19T05:24:13+00:00 1 Answers 12 views 0

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  1. minhkhang
    0
    2021-07-19T05:25:27+00:00
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    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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