If 2.54 L of hydrogen reacts with 3.13 L of oxygen at temperatures above 100 degrees Celsius, how many L of H2O(g) will form?

2 Answers
May 20, 2017

Answer:

Hydrogen gas is the limiting reactant, so the given reaction and volumes will produce 2.54 L of water.

Explanation:

Start with a balanced equation.

#"2H"_2("g")" + O"_2("g")"##rarr##"2H"_2"O"("g")#

This is a stoichiometry gas problem. Since the reactants and product are identified as liters (volume), we can use the relationships between the liters as we would do with moles.

This also a limiting reactant problem. You need to determine the volume of water that can be produced from the volumes given for each reactant gas. The gas that produces the least amount of water is the limiting reactant, and the amount of water that can be by the reaction is determined by the limiting reactant.

#color(blue)("Determine the volume of water produced by hydrogen gas"#.

The ratio between #"H"_2"# and #"H"_2"O"# is #2:2#, which is equal to #1:1#.

#2.54color(red)cancel(color(black)("L H"_2))xx(1"L H"_2"O")/(1color(red)cancel(color(black)("L H"_2)))="2.54 L H"_2"O"#

#color(blue)("Determine the volume of water produced by oxygen gas"#.

The ratio between the volume of #"O"_2"# and #"H"_2"O"# is #1:2#.

#3.13color(red)cancel(color(black)("L O"_2))xx(2"L H"_2"O")/(1color(red)cancel(color(black)("L O"_2)))="6.26 L H"_2"O"#

The limiting reactant is #"H"_2"#, which will produce #"2.54 L H"_2"O"#.

May 20, 2017

This problem involves determining the limiting reactant and using it to calculate the volume of #H_2O (g)# that forms, using the reaction

#2H_2(g) + O_2(g) rarr 2H_2O(g)#

We're not given a specific temperature or pressure, so we'll (albeit impractically) ignore the fact that the temperature is above #100 ^oC# for now.

To calculate the limiting reactant, we'll convert the volumes of the reactant gases to moles, and we'll assume that one mole of a gas occupies #22.4L#, since we're assuming standard temperature and pressure conditions. We'll need to divide the calculated moles for #H_2# by #2# to compensate the fact that there is a coefficient of #2# in front of it in the chemical equation:

#2.54LH_2((1molH_2)/(22.4LH_2)) = (0.113molH_2)/2 = 0.0567molH_2#

#3.13LO_2((1molO_2)/(22.4LO_2)) = 0.140molO_2#

It is clear that #H_2# is the limiting reactant, as it is present in the smaller amount.

We'll use the calculated value of #0.113molH_2# to find the moles, and then the volume in liters of the #H_2O(g)# that will form if the reaction is complete:

#0.113molH_2((2molH_2O)/(2 molH_2))((22.4LH_2O)/(1molH_2O)) = 2.54 L H_2O#

You'll notice this is the same value as the volume of #H_2# originally present, because the coefficients in front of #H_2# and #H_2O# are equal.