Anthony T. answered • 06/04/21

Patient Math & Science Tutor

Hi, Mandy I will try to teach you how to do these problems.

Consider the hypothetical reaction 2 A ====== > 3 B where A and B are hypothetic substances. A has a molecular mass of 25 and B 40.

Case 1: you are given the number of moles of A and asked to determine the number of moles of B produced.

Set up the conversion factor 3 moles B / 2 moles A which you get from the balanced equation. Note that the thing you want to find is in the numerator. Now multiply the factor by the number of moles of A you were given, let's say it's 5 moles of A. The calculation is 3 moles B / 2 moles A x 5 moles A = 7.5 moles of B produced. Note that moles of A units cancel leaving only moles B.

Case 2: you are given the # grams of A and asked to calculate the number of moles of B produced.

Set up this conversion factor 3 moles B / 2 x molecular mass of A (25). This becomes 3 moles B / 2 x 25 g A

or 3 moles of B / 50 g A. Again notice that the thing you are trying to calculate is in the numerator. Now multiply the conversion factor by the number of grams of A you are given, let's say 35 g.

The calculation is 3 moles B / 50 g A x 35 g A = 2.1 moles B produced.

Case3: you are given the number of grams of A and asked to find the number of grams of B produced.

The conversion factor here is 3 x molecular mass of B ( 3 x 40g) / 2 x molecular mass of A (2 x25g) or

3x 40g / 2 x 25 g = 2.4 g B / g A. Now multiply by the number of grams of A given ( let's say 5.0 g A) to get

2.4 g B / g A x 5.0 g A = 12.0 g B produced.

There are variations to these cases, but the general procedure is to write a conversion factor for what you are trying to find, putting the thing you are trying to find in the numerator, then multiply by the given quantity.

Now let's do the first problem. The problem asks for the moles of H2O produced from 163.6 g HNO3. The conversion factor is 2 moles H2O / 6 x mol. mass of HNO3 or 2 moles H2O / 6 x 63 g/mol but you are given 163.6 g HNO3, so the calculation becomes 2 mol H2O / 378 g HNO3 x 163.6 g HNO3 = 0.9 moles H2O produced.

This is probably not how this is presented in school, but I find it less confusing once you understand the procedure. Try the second problem yourself. The answer should be 0.9 moles H2O