Question

Question #afbfc

Answer 1

`"4.03 g"`

Explanation:

There are a few slightly different approaches that you can use here, but I'll show you how to answer this question by going to the number of moles of hydrogen first.

You know that every `1` mole of sulfuric acid contains `color(darkorange)(2)` moles of hydrogen. This means that your sample will contain twice as many moles of hydrogen as the number of moles of sulfuric acid.

Sulfuric acid has a molar mass of `color(blue)("98.079 g mol"^(-1)`, which tells you that `1` mole of sulfuric acid has a mass of `"98.079 g"`.

This means that your sample will contain

`196 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"SO"_4)/(98.079color(red)(cancel(color(black)("g")))) =(196/color(blue)(98.079))color(white)(.)"moles H"_2"SO"_4`

You can thus say that your sample contains

`(196/color(blue)(98.079)) color(red)(cancel(color(black)("moles H"_2"SO"_4))) * (color(darkorange)(2)color(white)(.)"moles H")/(1color(red)(cancel(color(black)("mole H"_2"SO"_4)))) = (color(darkorange)(2)/color(blue)(98.079) * 196)color(white)(.)"moles H"`

Finally, you know that atomic hydrogen has a molar mass of `color(purple)("1.00794 g mol"^(-1))`, which means that `1` mole of atomic hydrogen has a mass of `"1.00794 g"`.

You can thus say that your sample contains

`(color(darkorange)(2)/color(blue)(98.079) * 196) color(red)(cancel(color(black)("moles H"))) * "1.00794 g"/(1color(red)(cancel(color(black)("mole H")))) = ((color(darkorange)(2) * color(purple)(1.00794))/color(blue)(98.079) * 196)color(white)(.)"g"`

of hydrogen, which is equal to

`color(darkgreen)(ul(color(black)("mass of H = 4.03 g")))`

The answer is rounded to three sig figs, the number of sig figs you have for the mass of sulfuric acid.