Stoichiometric Problems

Hello There,

Question1:

1. Let's find the MW of NH3, 14g/mol +3g/mol = 17 g/mol

2. Find the Number of moles of NH3= 22.7 g * mol/17g = 1.34 mol of NH3

3. If 1.34 mol of NH3 is what we have, then we would need 1.34/2 mol (0.67 mol) of H2SO4.

4. Find MW of H2SO4 = 2g/mol +32g/mol + 16g/mol *4 = 98 g/mol

5. How many moles of H2SO4 do we have from 54.8 g?

54.8g *mol/98g = 0.559 mol of H2SO4 is available.

Well, the amount of H2SO4 will definitely be less than 0.67 mol which is lower than the needed number of moles to react ALL of the NH3 available. What does this mean? It means that H2SO4 is the limiting reagent. Therefore, you can only make the reaction go as far as the amount of available H2SO4 allows you.

3. Since we have 0.559 mol of H2SO4, we can use 0.559 *2 moles ( 1.118 moles) of NH3.

4. Now the product: the ration of H2SO4 to (NH4)2SO4 is 1:1 therefore, with 0.559 mol of H2SO4, 0.559 mol of (NH4)2SO4 is made

5. How many grams is that?
MW of (NH4)2SO4 is : [(14 +4) *2 + 32 +16*4] g/mol = 132 g/mol

Grams of (NH4)2SO4 is " 0.559 mol * 132 g/mol = 73.8 g of (NH4)2SO4 is made.

This was a limiting reagent problem!

Question 2:

1. Assume mass is 100g then the percentages are the same as the masses: 54.53g * mol/12g C: 9.15 g * mol/g H : 36.32 g *mol/16 g O