Question

Calculate the amount of water formed when `"8 g"` of `"H"_2"` is allowed to react completely with oxygen?

Answer 1

`~~"70 g H"_2"O"` will produced from `"8 g H"_2"`.

Explanation:

Balanced equation

`"2H"_2 + "O"_2"` `rarr` `"2H"_2"O"`

`color(red)(1.` Calculate mol `"H"_2"` by dividing its given mass by its molar mass `("2 g/mol")`. Do this by multiplying by the inverse of the molar mass.

`color(blue)(2.` Calculate mol `"H"_2"O"` by multiplying mol `"H"_2"` by the mol ratio between `"H"_2"O"` and `"H"_2"` from the balanced equation, with `"H"_2"O"` in the numerator.

`color(green)(3.` Calculate mass `"H"_2"O"` by multiplying by the molar mass of `"H"_2"O"` `("18 g H"_2"O")`.

`color(red)8color(black)cancel(color(red)("g H"_2))xx(color(red)(1color(black)cancel(color(red)("mol H"_2))))/(color(black)cancel(color(red)(2"g H"_2)))xx(color(blue)2color(black)cancel(color(blue)("mol H"_2"O")))/(color(blue)2color(black)cancel(color(blue)("mol H"_2)))xx(color(green)18color(green)("g H"_2"O"))/(color(green)1color(black)cancel(color(green)("mol H"_2"O")))="70 g H"_2"O"` (rounded to one significant figure)

Answer 2

72 g

Explanation:

The equation for the reaction of hydrogen gas and oxygen gas to form water is: `2H_2(g) + O_2(g) -> 2H_2O (l)`

So for every mole of `H_2` you will form a mole of water, assuming you have sufficient oxygen to react with it all.

The atomic weight of hydrogen atoms is 1, but the molar mass of diatomic `H_2` is 2 g/mol. Therefore, 8 g of `H_2` is 4 moles. Therefore you will form 4 moles of water. The molar mass of water is 18 g/mol so you will form 4 x 18 = 72 g of water.