When hydrogen reacts with oxygen under standard conditions, 10 liters of hydrogen are mixed with 2 liters of oxygen. Calculate the mass of the resulting water.

First you have to calculate the quantities of the two gases from the volume with the aid of the ideal gas equation:

The following shall apply:

p = pressure in Pascal (I assume normal pressure, 1013 hPa, i.e. 101300 Pa)

V = volume in cubic meters (indicated, 10 litres and 2 litres)

n = quantity of substance in mol

R = Ideal gas constant (8,314 J/mol*K)

T=temperature in Kelvin (I proceed from normal temperature, 0 degrees Celsius, i.e. 273,15 K)

Use then gives:

Now we need the reaction equation:

It can be seen at the reaction equation that the ratio of the substance quantity between hydrogen and oxygen is 2:1.

In the event that all hydrogen, i.e. 0.446 mol, reacts, the amount of oxygen required for this would be:

However, we have only 0.089 mol of oxygen present. This is completely reacted, for which it requires the following amount of hydrogen:

Thus, only about 0.178 mol of hydrogen reacts from the 0.446 mol of hydrogen, since too little oxygen is present for a complete reaction of the 0.446 mol of hydrogen (only 0.089 mol, although 0.223 mol would be required).

It is also apparent from the reaction equation that the proportion of hydrogen and water in the quantity is 1:1 (actually 2:2, which is the same as 1:1).
For the quantity of water formed, the following applies:

Thus, 0.178 mol of water are formed. With the molar mass of water, the mass of the water formed can now be calculated:

The mass of the resulting water is therefore 3.207 grams.

Note:

The quantities of substances of the two gases can also be calculated via the molar volume. In addition, almost the same results should be obtained with any deviation in the 3rd or 4th decimal place. This is enough for most calculations.
Personally, I prefer the ideal gas equation, because there is also temperature and pressure.