If 50.0 dm3 of methane, ch4, react with 10.0 dm3 of air, calculate the grams of water produced.
ch4 (g) + 2 o2 (g) --> co2 (g) + 2 h2o (l)

Answer : The mass of produced will be, 8.1 grams.

Explanation : Given,

Volume of =    

Volume of =

Molar mass of = 18 g/mole

First we have to calculate the moles of and .

As, 22.4 L volume of present in 1 mole of

So, 50 L volume of present in mole of

and,

As, 22.4 L volume of present in 1 mole of

So, 10 L volume of present in mole of

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

From the balanced reaction we conclude that

As, 2 moles of react with 1 mole of

So, 0.45 moles of react with moles of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of .

As, 2 moles of react to give 2 moles of

As, 0.45 moles of react to give 0.45 moles of

Now we have to calculate the mass of .

Therefore, the mass of produced will be, 8.1 grams.