Question

A sample of PH3, occupying 1.50 L at 25 oC and 0.80 atm, reacts with 0.080 atm, reacts with 2.28L of O2 at 20.0 oC and 0.100 atm according to the following equation:

4 PH3(g) + 8 O2(g) => P4O10(s) + 6 H2O(g)

How many grams of each product is produced? How many grams of each reactant are left at the end of the reaction? What is the final pressure at the end of the reaction? (The final volume is 3.88L and the final temperature is 22.5C).

Answer 1

The ideal gas equation can be used to obtain the pressure, number of moles, volume and temperature of a gas.

What is the ideal gas equation?

The ideal gas equation gives the relationship between pressure, number of moles, volume and temperature.

To obtain the number of moles of PH3

PV = nRT

n = PV/RT

n = 0.80 atm * 1.50 L/0.082 atmL/Kmol * 298 K

n = 0.049 moles

Number of moles of O2

PV = nRT

n = PV/RT

n = 0.100 atm *  2.28L /0.082 atmL/Kmol * 293 K

n = 0.0095 moles

Fom the reaction equation;

4 moles of PH3 reacts with 8 moles of O2

0.049 moles of PH3 reacts with  0.049 moles * 8/4 moles

= 0.098

Hence O2 is the limiting reacatant.

Mass of P4O10 produced =

8 moles of O2 yields 1 mole of P4O10

0.0095 moles  of O2 yields 0.0095 moles  * 1 mole/8 moles

= 0.0012 moles

Mass of P4O10  = 0.0012 moles * 284 = 0.34 g

Mass of water;

8 moles of O2 yields 6mole of water

0.0095 moles  of O2 yields 0.0095 moles * 6 moles/8 moles

=0.0071 moles

Mass of water = 0.0071 moles * 18 g/mol = 0.13 g

Since 4 moles of PH3 reacts with 8 moles of O2

x moles of PH3 reacts with 0.0095 moles  of O2

x = 0.0048 moles

Moles of PH3 left over = 0.049 moles - 0.0048 moles = 0.0442 moles

Mass of PH3 left over = 0.0442 moles * 34 g/mol

= 1.5 g

No mass of O2 is left over because the limiting reactant is used up.

Total moles of gas left after the reaction = 0.0442 moles +  0.0071 moles = 0.0513 moles

P = nRT/V

P = 0.0513 moles * 0.082 atmL/Kmol * 295.5/3.88L

P= 0.32 atm'

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