Help with Chem rate of reaction problem

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SUMMARY

The discussion focuses on calculating the rates of production of P4 and H2 from the reaction 4PH3(g) ------- P4(g) + 6H2(g). Given that 0.0063 mol of PH3 is consumed per second in a 2.0 L container, stoichiometric relationships are used to determine the rates of formation for P4 and H2. Specifically, for every 4 moles of PH3 consumed, 1 mole of P4 and 6 moles of H2 are produced. Therefore, the rate of production of P4 is 0.001575 mol/s and the rate of production of H2 is 0.00945 mol/s.

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pata320
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4PH3(g) ------- P4(g) + 6H2(g)
If, in a certain experiment, over a specific time period, 0.0063 mol PH3 is consumed in a 2.0 L container each second of reaction, what are the rates of production of P4 and H2 in this experiment.

Any idea about this problem? divide 2.0 to 1.0 and then .0063 to .00315 but from there I have no clue. Anybody have a clue? Thanks a bunch!
 
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pata320 said:
4PH3(g) ------- P4(g) + 6H2(g)
If, in a certain experiment, over a specific time period, 0.0063 mol PH3 is consumed in a 2.0 L container each second of reaction, what are the rates of production of P4 and H2 in this experiment.

the rate of consuming of PH3 is 0,0063 mol / s. Now use stoichiometry to get to know the rates of formation of P4 and H2:

1. when 4 mol PH3 has reacted, then 1 mol P4 is formed --> How much P4 is formed per second when 0,0063 mol PH3 has reacted (per second) ?

2. when 4 mol PH3 has reacted, then 6 mol H2 is formed --> How much H2 is formed per second when 0,0063 mol PH3 has reacted (per second) ?
 

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