18. Hydrogen gas reacts with oxygen gas to form water vapor. If 34.5 g of hydrog

Question

18. Hydrogen gas reacts with oxygen gas to form water vapor. If 34.5 g of hydrogen gas react with 26.6 g of oxygen gas, how many grams of water vapor will be produced? 2H2(g) + O2(g) --> 2H2O(g)

20. The reaction of 22.5 g of Fe with excess of oxygen gas produced 24.2 g of Fe2O3. Calculate the percent yield.

21. Milk of magnesia contains magnesium hydroxide. Mg(OH)2 reacted with HCl to form aqueous magnesium chloride and liquid water. Write the properly balanced equation for this reaction.

22. Lithium reacted with oxygen gas to form lithium chloride. The chemical formula for lithium chloride is___________.

(A). LiO (B). LiO2 (C). Li2O3 (D). Li2O (E). Li2O4

23. In the following reaction, how many grams of sodium sulfate are needed to react with 3.08 moles of strontium nitrate? (Hint: stoichiometry mol Na2SO4: mol Sr(NO3)2 = 1 mol: 1 mol) Sr(NO3)2 (aq) + Na2SO4 (aq) --> SrSO4 (s) + 2 NaNO3 (aq) (balanced?)

24. Which of the following represents a chemical property of oxygen gas?

(A). It is gaseous at room temperature. (B). It reacts explosively with propane. (C). It is less dense than air. (D). It is colorless

Explanation / Answer

18. 2H2 + O2 -> 2H2O

Turning reactants to moles:

34.5 grams of H2 * (1 mol / 2 grams) = 17.25 moles of H2

26.6 grams fo O2 * (1mol / 32 grams) = 0.83125 moles of O2

As we can see, limiting reactant is oxygen. and grams of product will be:

0.83125 moles of O2 * (2 moles of H2O / 1 mol of O2) * (18 grams/mol of H2O) = 29.925 grams of H2O

20. % yield = actual grams / theoretical grams

actual grams = 24.2 of Fe2O3

Theoretical grams:

4Fe + 3O2 -> 2Fe2O3

22.5 g of Fe * (1 mol / 55.845 g) = 0.4029 moles of Fe

0.4029 moles of Fe * (2 moles of Fe2O3 / 4 moles of Fe) = 0.20145 moles of Fe2O3

Now to grams -> 0.20145 moles of Fe2O3 * (160 g / mol) = 32.23 grams

% yield = 24.2 grams / 32.23 grams = 0.7508

21. Mg(OH)2 + 2HCl -> MgCl2 + 2H2O

22. Assuming it is lithium oxide instead of chloride, answer is Li2O

23. Sr(NO3)2 (aq) + Na2SO4 (aq) --> SrSO4 (s) + 2NaNO3 (aq) (BALANCED)

3.08 grams of Sr(NO3)2 * (1 mol / 211.63 grams) = 0.01455 moles of Sr(NO3)2

As we have a 1 on 1 stoichiometric relation:

0.01455 moles of Na2SO4 * (142.04 g / mol) = 2.0672 grams

24. It reacts explosively with propane, it is the only option regarding a chemical change.

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