P4(s) + H2(g) -----> PH3(g) Calculate the mass of hydrogen required to combine w

Question

P4(s) + H2(g) -----> PH3(g)

Calculate the mass of hydrogen required to combine with 71.19 g of phosphorus and Calculate the mass of PH3 produced when 71.19 g of phosphorus combines with the amount of hydrogen calculated in part b.

How do you go about this? Please give detailed answer.

Thanks

Explanation / Answer

P4 mol wt = 30.97x4 = 123.88 71.19 means 71.19/123.88 =0.5746 moles P4 + 6H2 ---> 4PH3 0.5746 moles P4 requires 0.5746 x6 moles of H2 = 0.5746 x12 x2 =13.79 gms of H2 0.5746 moles P4 gives 4 x0.5746 = 2.298 moles of PH3 = 2.298 x33.97 = 78.06 gm PH3 hence mass of H2 required =13.79 gm mass of PH3 produced = 78.06 gm

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