Chapter 6

Lead (IV) chloride reacts with fluorine gas to give lead (IV) fluoride and \(\mathrm{Cl}_{2}\). If \(0.023\) moles of fluorine gas reacts with \(5.3\) grams of lead (IV) chloride, what mass of lead (IV) fluoride will be formed?

a. Iron (III) oxide reacts with hydrogen gas to form elemental iron and water, according to the balanced equation shown below. How many moles of iron will be formed from the reduction of excess iron (III) oxide by \(0.58\) moles of hydrogen gas? b. When an impure sample containing an unknown amount of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is reacted with excess hydrogen gas, \(0.16\) moles of solid Fe are formed. How many moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) were in the original sample?

The Harber process is used making ammonia from nitrogen and hydrogen according to the equation shown below. The yield of the reaction, however, is not \(100 \%\). a. Suppose you end up with \(6.2\) moles of ammonia, but the reaction stoichiometry predicts that you should have \(170.0\) grams of ammonia. What is the percent yield for this reaction? b. If you started with \(6.2\) grams of nitrogen and you produce \(6.2\) grams of ammonia what would be the percent yield?

Ammonia is produced industrially from nitrogen and hydrogen according to the equation: $$ \mathrm{N}_{2}+3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3} $$ a. If you are given \(6.2\) moles of nitrogen how many mole of ammonia could you produce? b. How many moles of hydrogen would you need to fully react with \(6.2\) moles of nitrogen? C. If you wished to produce 11 moles of ammonia how many moles of nitrogen would you need to start with?

Aluminum and chlorine gas react to form aluminum chloride according to the balanced equation shown in below. $$ 2 \mathrm{Al}(s)+3 \mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{AlCl}_{3}(s) $$ If \(17.467\) grams of chlorine gas are allowed to react with excess \(\mathrm{Al}\), what mass of solid aluminum chloride will be formed?