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Question

4 PH3(g) P4(g) + 6 H2(g)
If, in a certain experiment, over a specific time period, 0.0045 mol PH3 is consumed in a 1.8 L container each second of reaction, what are the rates of production of P4 and H2 in this experiment? The answers are in mol/L.s

Answer 1

4PH3(g) ===> P4(g) + 6H2(g)

d[PH3]/dt = 0.0045molPH3/1.8L-s = 0.0025mol/L-sec

2.5x10^-3molPH3/L-sec x 1molP4/4molPH3 = 6.25 x 10^-4 mole P$ per sec

2.5x10^-3molPH3/L-sec x 6molH2/4molPH3 = 3.75 x 10^-3 mole H2 per sec