Stoichiometric Solutions

Question

What is the step-by-step procedure solution and answers to these one?

How many moles CO2 of is produced by the complete combustion of 58 g of butane (C4H10)?

2C4H10 + 13O2 → 8CO2 +10H2O2.

If 5.0 moles of Al react with excess O2, how many moles of Al2O3 can be formed?

4Al + 3O2 → 2 Al2O33.

How many L of barium sulfate (BaSO4 ) will be produces from a reaction of 32 mL of 0.310 M barium nitrate and 28 mL of 0.380 M aluminum sulfate?

3 Ba(NO3)2 (aq) + Al2(SO4)3 (aq) → 3BaSO4 (s) + 2Al(NO3)3 (aq)

Answer 1

For the first one:

1. Find the molar mass of C4H10 
2. C = 12g x 4 = 48g H = 1g x 10 = 10g; molar mass of C4H10 = 58 g/mol
3. Convert grams of C4H10 into moles C4H10 
4. 58g C4H10 x (1 mol / 58g) = 1 mol C4H10
5. Use the mole ratio to convert moles C4H10 to moles CO2 [look at the coefficients in the equation to determine the mole ratio]
6. Mole ratio: 2 mol C4H10 to 8 mol CO2
7. 1 mol C4H10 x (8 mol CO2 / 2 mol C4H10) = 4 mol CO2

For the second one:

1. Use the mole ratio to convert moles Al to moles Al2O3 [like step 3 from the previous problem]. That’s it!

For the third one:

You will have to do 2 separate calculations to find the limiting reagent [Ba(NO3)2 or Al2(SO4)3]. The limiting reagent is the one that produces the smaller number of moles of BaSO4. The smaller number of moles of BaSO4 will be your answer.

1. Find the number of moles of Ba(NO3)2 and Al2(SO4)3 using M = mol/L
2. 0.310 M = mol Ba(NO3)2 / 0.032 L Ba(NO3)2 = 0.00992 mol Ba(NO3)2
3. Repeat process for Al2(SO4)3 [be sure to convert mL -> L first]
4. Use the mole ratio to convert moles Ba(NO3)2 and moles Al2(SO4)3 to moles BaSO4
5. 0.00992 mol Ba(NO3)2 x (3 mol BaSO4 / 3 mol Ba(NO3)2) = 0.00992 mol BaS04
6. Repeat process for Al2(SO4)3 
7. Compare the two quantities. The smaller number of moles of BaSO4 will be your answer.