If 3.1gm Phosphorous is present in a sample of Ca3(PO2)4, what is the weight of oxygen in the sample
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Answer:
3.199gram molecular weight.
Explanation:
weight of Phosphorous given=3.1gm
Molecular weight of Ca3(PO2)4=120+124+128=372 gm
124 gm P is present in 372 gm of the compound, hence amount of sample provided= (372/124)x3.1= 3x3.1= 9.3gm
Again, 372gm of Ca3(PO2)4 contain 128 gm of O, therefore, amount of O in the sample= (128/372)x9.3= 0.344086x9.3= 3.19999gm of Oxygen.
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