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.

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.