Limiting and Excess Stoichiometry Questions

Question

Please Help. Thanks alot.

1. What mass of Al2 SO4 is needed to create 100g of Al(NO3 )3 in the reaction below (NOT BALANCED):

Al2(SO4)3 + Ba(NO3) =

2. Copper (II) chloride reacts with sodium nitrate in a double displacement reaction. Assume this reaction takes place on a planet with different solubility rules than ours and the reaction DOES occur!

a) Write the balanced equation for the reaction given above:

b) If 15 grams of copper (II) chloride react with 20 grams of sodium nitrate, how much sodium chloride can be formed?

c) What is the limiting reagent for the reaction in #2? _________

d) How many grams of copper(II) nitrate is formed?

e) How much of the excess reagent is left over (in grams) in this reaction?

Answer 1

To start off, the formula shown for the barium nitrate is incorrect: it should be Ba(NO₃)₂. Now we can begin the problem.

1) Al₂(SO₄)₃ + 3 Ba(NO₃)₂ ---> BaSO₄ + 2 Al(NO₃)₃

Since we are given a mass and want to find a mass, we will need the molar masses of the two substances. Al₂(SO₄) = 310.1 g/mol & the Al(NO₃)₃ = 213.0 g/mol

(310.1 g Al₂(SO₄)₃/1 mol Al₂(SO₄)₃) (1 mol Al₂(SO₄)₃)/2 mol Al(NO₃)₃) (1 mol Al(NO₃)₃/213.0 g Al(NO₃)₃) (100 g Al(NO₃)₃) = 72.79 g Al₂(SO₄)₃ You have a max of 3 sig figs given, so you should record the answer as ~72.8 g. {Note that the number 100 could have as few as 1 sig fig and is poorly communicated as the number 100}

2)

a) CuCl₂ + 2 NaNO₃ ---> Cu(NO₃)₂ + 2 NaCl

b) Here, you have two different quantities of the reactants, so you must calculate which will give the least amount of NaCl. That would be the correct theoretical yield & the reactant that gave that would be the limiting reactant, while the other one would be in excess.

i) Using the given amount of CuCl₂ solve for the mass of NaCl that would be formed:

(58.5 g NaCl/1 mol NaCl) (2 mol NaCl/1 mol CuCl₂) (1 mol CuCl₂/135.4 g CuCl₂) 15 g CuCl₂ = 12.96 g = 13 g NaCl.

ii) Using the given amount of sodium nitrate, do the same thing:

(58.5 g NaCl/1 mol NaCl) (2 mol NaCl/2 mol NaNO₃) (1 mol NaNO₃/85 g NaNO₃) 20 g NaNO₃ = 3.98 g = 4.0 g NaCl

Since we get less NaCl from the sodium nitrate, the sodium nitrate is the limiting reactant (answer to c) and the yield is ~ 4.0 g NaCl (b)

d) To find the mass of copper (II) nitrate, do another mass/mass calculation using the amount of sodium nitrate, since we now know that's the LR:

(188 g Cu(NO₃)₂/1 mol Cu(NO₃)₂) (1 mol Cu(NO₃)₂/2 mol NaNO₃) (1 mol NaNO₃/85 g NaNO₃) 20 g NaNO₃ = 22 g Cu(NO₃)₂

e) To find the amount of excess CuCl₂ (the excess reactant) that is left over, we calculate how much of it was reacted (using the given NaNO₃ again, and subtract that from the 15 g we had:

(135.4 g CuCl₂/1 mol CuCl₂) (1 mol CuCl₂/2 mol NaNO₃) (1 mol NaNO₃/85 g NaNO₃) 20 g NaNO₃ = 15.9 g = ~16 g CuCl₂ that reacted.

So, 20 g - 16 g = 4 g CuCl₂ is left

Answer 2

(1.) First, let's write the balanced equation:

Al₂(SO₄)₃ + 3Ba(NO₃)₂ ==> 2Al(NO₃)₃ + 3BaSO₄ ... balanced equation

100 g Al(NO₃)₃ x 213 g/mol x 1 mol Al₂(SO₄)₃ / 2 mol Al(NO₃)₃ x 342 g/mol = 80.3 g Al₂(SO₄)₃ needed. If you refer to 100 g starting material as having only 1 significant figure, the answer would be 80 g.

(2.) CuCl₂ + 2NaNO₃ ==> Cu(NO₃)₂ + 2NaCl ... balanced equation

b) 15 g CuCl₂ x 1 mol CuCl₂/134 g x 2 mol NaCl/mol CuCl₂ x 58 g/mol NaCl = 12.99 g NaCl formed

20 g NaNO₃ x 1 mol NaNO₃/85 g x 2 mol NaCl/2 mol NaNO₃ x 58 g/mol NaCl = 13.65 g NaCl formed

Thus, 12.99 g = 13 g NaCl can be formed

c) Limiting reagent is CuCl₂ (see above calculations)

d) 15 g CuCl₂ x 1 mol CuCl₂/134 g x 1 mol Cu(NO₃)₂ / 1 mol CuCl₂ x 188 g/mol Cu(NO₃)₂ = 21 g Cu(NO₃)₂

e) Excess reagent is NaNO₃; amount NaNO₃ used up = 15 g CuCl₂ x 1 mol/134 g x 2 mol NaNO₃/mol CuCl₂ = 0.224 moles used up. Moles initially present = 20 g x 1 mol/85 g = 0.235 moles

Moles NaNO₃ remaining = 0.235 - 0.224 = 0.011 moles

mass NaNO₃ remaining = 0.011 moles x 85 g/mol = 0.94 g NaNO₃ left over