Question

Question #8d65b

Answer 1

Here's what I got.

Explanation:

Since you didn't provide information about the conditions for pressure * and temperature* at which the gases are kept, I will assume that you're working at Standard Temperature and Pressure, STP.

STP conditions are defined as a pressure of `"100 kPa"` and a temperature of `0^@"C"`. Under these conditions, one mole of any ideal gas occupies `"22.7 L"` - this is known as the molar volume of a gas at STP**.

Now, an important thing to remember when dealing with a reaction that involves gases kept under the same conditions for pressure and temperature is that the mole ratio that exists between the gases is equivalent to the volume ratio.

Your balanced chemical equation looks like this

`color(red)(2)"H"_text(2(g]) + "O"_text(2(g]) -> 2"H"_2"O"_text((g])`

Notice the `color(red)(2):1` mole ratio that exists between hydrogen gas and oxygen gas. This tells you that the reaction consumes `color(red)(2)` moles of hydrogen gas for every `1` mole of oxygen gas.

Since all three gases are under the same conditions for pressure and temperature, you can say that the reaction requires a volume of hydrogen as that is twice as large as the volume of oxygen gas.

You know that you're using `"6.0 L"` of hydrogen gas, which means that the reaction will require

`6.0 color(red)(cancel(color(black)("L H"_2))) * overbrace("1 L O"_2/(color(red)(2)color(red)(cancel(color(black)("L H"_2)))))^(color(purple)("2:1 volume ratio")) = color(green)(|bar(ul(color(white)(a/a)"3.0 L O"_2color(white)(a/a)|)))`

So, in order for the reaction to consume all the hydrogen gas, you need `"2.0 L"` of oxygen gas present.

If you want to convert this to grams of oxygen gas, use the molar volume of a gas to determine how many moles of oxygen would occupy this volume at STP

`3.0color(red)(cancel(color(black)("L O"_2))) * overbrace("1 mole O"_2/(22.7color(red)(cancel(color(black)("L O"_2)))))^(color(brown)("molar volume of a gas at STP")) = "0.1322 moles O"_2`

Finally, use the molar mass of oxygen gas to figure out how many grams would contain this many moles

`0.1322 color(red)(cancel(color(black)("moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = color(green)(|bar(ul(color(white)(a/a)"4.2 g O"_2color(white)(a/a)|)))`

The answers are rounded to two sig figs.

SIDE NOTE Many textbooks and online sources still list STP conditions as being a pressure of `"1 atm"` and a temperature of `"0^@"C"`.

Under these conditions for pressure and temperature, the molar volume of a gas at STP is equal to `"22.4 L"`.

If those are the values given to you, redo the calculations by replacing `"22.7 L"` with `"22.4 L"`.