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

1 Answer
Trevor Ryan.
Jan 17, 2016

#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#.

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