Question

Please can I get help on these?

16. When 1.5 L of methane gas combust in an excess of oxygen at standard temperature and pressure, how many liters of water vapor will be produced?
CH4 (g) + 2O2 (g) yields CO2 (g) + 2H2O (g) (5 points)
Answer

A) 0.75 L

B) 1.5 L

C) 3.0 L

D) 6.0 L


Question 17

17. When 4.50 L of hydrogen gas react with an excess of nitrogen gas at standard temperature and pressure, how many liters of ammonia gas will be produced?
N2 (g) + 3H2 (g) yields 2NH3 (g) (2 points)
Answ

Answer 1

the answer for the first question.

\[CH _{4}+ 2O _{2} = CO _{2}+ 2H _{2}O\]

the temperature and pressure remains the same in the given question.
so, at constant Temp. and Pressure, volume is directly proportional to number of moles.

and as,
\[\frac{ vol. of methane combusted }{ vol. of water vapour produced }= \frac{ no. of moles of methane }{ no. of moles of water vapour. }\]
and vol. of methane combusted = 1.5L
vol. of water vapour produced = we need to find out = say x

no. of moles of methane combusted = 1
no. of moles of water vapour produced = 2

putting all these things in the above relation.
we get,
the volume of water vapour produced = 3.0L

Answer 2

same procedure for the next question.

now do it yourself on the same lines.

the answer will come out to be 3.0L in this question too....