# How many grams of water could be made from 5.0 mol H_2 and 3.0 mol O_2?

Mar 27, 2017

You need a stoichiometric equation to represent the formation of water:

#### Explanation:

${H}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow {H}_{2} O \left(l\right)$

Given the stoichiometry, dihydrogen is in deficiency, and ONLY $2.5 \cdot m o l$ dioxygen will react to give $5 \cdot m o l$ water.

And this represents a mass of $5 \cdot \cancel{m o l} \times 18.01 \cdot g \cdot \cancel{m o {l}^{-} 1} = 90 \cdot g$.

What mas of dioxygen will remain after reaction?

Mar 27, 2017

90 g

#### Explanation:

The balanced equation is: $2 {H}_{2} \left(g\right) + {O}_{2} \left(g\right) \to 2 {H}_{2} O \left(l\right)$

From this you can see that 5 moles of hydrogen gas would react with 2.5 of the available 3 moles of oxygen gas, to form 5 moles of water.

The molar mass of water is 18 g/mol, so you would end up with 5 x 18 = 90 g of water.