Asked by Name
Consider the following reaction in water:
P4 (s) + 6 H2O (l) → 2 PH3 (g) + 2 H3PO3 (aq)
If you allow 23.42 g of P4 to react what would be the pressure that the dry gas would exert on a 15.00
mL container at STP. Assume all of the phosphorus reacted.
This can be solved by finding the partial pressures of the reactants correct?
P4 (s) + 6 H2O (l) → 2 PH3 (g) + 2 H3PO3 (aq)
If you allow 23.42 g of P4 to react what would be the pressure that the dry gas would exert on a 15.00
mL container at STP. Assume all of the phosphorus reacted.
This can be solved by finding the partial pressures of the reactants correct?
Answers
Answered by
bobpursley
well, you get 2 moles of gas for each mole of P4.
moles P4=23.42g/4*30.97=.189moles
Moles gas=2*.189
PV=nRT
P=nRT/V=
solve for P.
moles P4=23.42g/4*30.97=.189moles
Moles gas=2*.189
PV=nRT
P=nRT/V=
solve for P.
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