Practice Problems Week 7

1. Find the oxidation number of chromium in each of the following compounds:

(a)

(b)

(c)

2. For each of the following equations, (a) complete and balance the equation; (b) determine if the reaction is redox or not; (c) for redox, write the oxidation and reduction half reactions and (d) indicate the element that is reduced, the element that is oxidized, the reducing agent and the oxidizing agent.

(i) sodium carbonate + iron(III) nitrate -> sodium nitrate + iron(III) carbonate

(a)

(b) It is not redox (all atoms retain their formal charges)

(ii) aluminum + nitric acid (hydrogen nitrate) -> aluminum nitrate + hydrogen gas

(a)

(b) It is redox because Al and H are changing charge states

(d) Oxidizing agent: nitric acid; reducing agent: aluminum

(iii) aluminum + bromine -> aluminum bromide

(a)

(b) It is redox because Al and Br change charge states

(d) Oxidizing agent: bromine; reducing agent: aluminum

3. Given the equation:

(a) How many moles of SO₂ are produced from 6.68 moles of O₂?

(b) How many grams of FeS₂ are required to react with 0.750 moles O₂?

(d) If 31.8 grams of SO₂ were isolated in problem 3(c) above, what is the percent yield of SO₂?

4. Given the reaction:

(a) What is the limiting reactant?

To determine which reactant is limiting, we need to see if is greater or less than :

So is clearly greater than and so hydrogen is in excess. That means N₂ will go away first and stop the reaction which makes it the limiting reactant. So we calculate the yield based on N₂.

(b) What is the theoretical yield of ammonia (NH₃) when 15.0 grams of N₂ is mixed with 5.00 grams of H₂?

5. Given the reaction:

(a) What is the theoretical yield when 20.0 grams of Mg are mixed with 10.0 grams of N₂?

So is less than and so N₂ is in excess and there isn't enough Mg. So Mg will go away first. We must calculate the theoretical yield based on Mg.

(b) What is the limiting reactant? Magnesium, because it goes away first, stopping the reaction.