Question

Question #c2d3e

Answer 1

`"440 g BaO"` are required to produce `"2.9 mol H"_2"O"`.

Explanation:

Balanced equation

`"BaO(s) + H"_2"SO"_4("aq")"` `rarr` `"BaSO"_4("s") + "H"_2"O(l)"`

The assumption is that `"H"_2"SO"_4"` is in excess. We can determine the moles `"BaO"` required by multiplying moles `"H"_2"O"` produced by the mole ratio between `"BaO"` and `"H"_2"O"` from the balanced equation with moles `"BaO"` in the numerator. This will cancel moles `"H"_2"O"` and leave moles `"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 `"2.9 mol BaO"`, multiply by its molar mass `("153.326 g/mol")`.

`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