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.

Oct 21, 2016

$9.65 \cdot {10}^{23} \text{molecules B}$

Explanation:

The only two substances that are of interest to you are $\text{A}$ and $\text{B}$, so you can write the chemical equation as

$\textcolor{red}{2} \text{A" + ... -> color(blue)(3)"B} + \ldots$

and assume that it is balanced.

Now, notice that the reaction consumes $\textcolor{red}{2}$ moles of substance $\text{A}$ and produces $\textcolor{b l u e}{3}$ moles of substance $\text{B}$.

This tells you that regardless of the number of moles of $\text{A}$ that take part in the reaction, you will always end up with $\frac{\textcolor{b l u e}{3}}{\textcolor{red}{2}}$ times more moles of $\text{B}$.

Use the molar mass of substance $\text{A}$ 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 $\text{B}$ 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 $\text{B}$, you must use Avogadro's constant, which tells you that one mole of a molecular substance contains exactly $6.022 \cdot {10}^{23}$ molecules of that substance.

In your case, the number of molecules of $\text{B}$ 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"))))

$= \textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{9.65 \cdot {10}^{23} \text{molecules B}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.