Question

How many grams of water can be prepared from 10.10 grams of hydrogen gas and excess oxygen gas at standard conditions?

Answer 1

`90.9g`

Explanation:

First write a balanced chemical equation for the reaction :

`2H_2+O_2->2H_2O`

Since oxygen is in excess, it implies that hydrogen is the limiting reagent and will decide how much product is formed, according to the mole ratio of the balanced equation.

`n_(H_2)=m/(M_r)=(10.10g)/(2g//mol)=5.05mol`.

`therefore moles of water formed will also be`5.05mol#.

`therefore m_(H_2O)=nxxM_r`

`=5.05xx(2+16)`

`=90.9g`.