Question

In a chemical reaction, exactly 2 mol of substance A react to produce exactly 3 mol of substance B. How many molecules of substance Bare produced when 28.3 g of substance A reacts? The molar mass of substance A is 26.5 g/mol.

Answer 1

`9.65 * 10^(23)"molecules B"`

Explanation:

The only two substances that are of interest to you are `"A"` and `"B"`, so you can write the chemical equation as

`color(red)(2)"A" + ... -> color(blue)(3)"B" + ...`

and assume that it is balanced.

Now, notice that the reaction consumes `color(red)(2)` moles of substance `"A"` and produces `color(blue)(3)` moles of substance `"B"`.

This tells you that regardless of the number of moles of `"A"` that take part in the reaction, you will always end up with `color(blue)(3)/color(red)(2)` times more moles of `"B"`.

Use the molar mass of substance `"A"` and the mass of the sample to figure out how many moles are taking part in the reaction

`28.3 color(red)(cancel(color(black)("g"))) * "1 mole A"/(26.5color(red)(cancel(color(black)("g")))) = "1.068 moles A"`

Now all you have to do is use the aforementioned mole ratio to calculate how many moles of `"B"` you'd get

`1.068 color(red)(cancel(color(black)("moles A"))) * (color(blue)(3)color(white)(a)"moles B")/(color(red)(2)color(red)(cancel(color(black)("moles A")))) = "1.602 moles B"`

Now, to find the number of molecules of `"B"`, you must use Avogadro's constant, which tells you that one mole of a molecular substance contains exactly `6.022 * 10^(23)` molecules of that substance.

In your case, the number of molecules of `"B"` produced by the reaction will be

`1.602 color(red)(cancel(color(black)("moles B"))) * (6.022 * 10^(23)"molecules B")/(1color(red)(cancel(color(black)("mole B"))))`

`= color(green)(bar(ul(|color(white)(a/a)color(black)(9.65 * 10^(23)"molecules B")color(white)(a/a)|)))`

The answer is rounded to three sig figs.