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.

Md Ariful Islam

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Md Ariful Islam

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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 .

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