# Question c2d3e

Nov 21, 2017

$\text{440 g BaO}$ are required to produce $\text{2.9 mol H"_2"O}$.

#### Explanation:

Balanced equation

$\text{BaO(s) + H"_2"SO"_4("aq")}$$\rightarrow$$\text{BaSO"_4("s") + "H"_2"O(l)}$

The assumption is that $\text{H"_2"SO"_4}$ is in excess. We can determine the moles $\text{BaO}$ required by multiplying moles $\text{H"_2"O}$ produced by the mole ratio between $\text{BaO}$ and $\text{H"_2"O}$ from the balanced equation with moles $\text{BaO}$ in the numerator. This will cancel moles $\text{H"_2"O}$ and leave moles $\text{BaO}$.

2.9color(red)cancel(color(black)("mol H"_2"O"))xx(1"mol BaO")/(1color(red)cancel(color(black)("mol H"_2"O")))="2.9 mol BaO"

To determine the mass of $\text{2.9 mol BaO}$, multiply by its molar mass $\left(\text{153.326 g/mol}\right)$.

2.9color(red)cancel(color(black)("mol BaO"))xx(153.326"g BaO")/(1color(red)cancel(color(black)("mol BaO")))="440 g BaO"# rounded to two significant figures