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To completely react 0.40 mol of magnesium hydroxide, 0.20 mol of hydrochloric acid will be required. Mg(OH) 2 (aq) + 3O 2 (g) 2H 2 O (l) MgCl 2 (aq) 0.40.

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Presentation on theme: "To completely react 0.40 mol of magnesium hydroxide, 0.20 mol of hydrochloric acid will be required. Mg(OH) 2 (aq) + 3O 2 (g) 2H 2 O (l) MgCl 2 (aq) 0.40."— Presentation transcript:

1 To completely react 0.40 mol of magnesium hydroxide, 0.20 mol of hydrochloric acid will be required. Mg(OH) 2 (aq) + 3O 2 (g) 2H 2 O (l) MgCl 2 (aq) 0.40 mol Mg(OH) 2 x 2 mol O 2 = O.80 mol KClO 3 1 mol Mg(OH) 2 FALSE! To completely react 0.40 mol of magnesium hydroxide, 0.80 mol of hydrochloric acid will be required. TRUE OR FALSE?

2 Before going to lab, a student read in her lab manual that the percent yield for a difficult reaction was likely to be only 40.% of the theoretical yield. The student's prelab stoichiometric calculations predict that the theoretical yield should be 12.5 g. What's the student's likely actual yield? 0.40 = Actual Yield 12.5 g Actual Yield = (0.40) (12.5g) = 5.0 g % Yield = Actual Yield x 100 Theoretical Yield

3 When elemental copper is placed in a solution of silver nitrate, an oxidation reduction reaction takes place which produces elemental silver. What mass of copper is required to remove all the silver from a silver nitrate solution containing 1.95 g of silver nitrate? (HINT... copper (II) nitrate is one of the products)

4 1.95 g AgNO 3 x 1 mol AgNO 3 x 1 mol Cu x g Cu g AgNO 3 2 mol AgNO 3 1 mol Cu = g Cu What mass of copper is required to remove all the silver from a silver nitrate solution containing 1.95 g of silver nitrate? Cu(s) + 2AgNO 3 (aq) 2Ag (s) + 3Cu(NO 3 ) 2 (aq) First, write the balanced equation... Next, do the conversion...

5 20.0 g N 2 H 4 x 1 mol N 2 H 4 x 1 mol N 2 = mol N g N 2 H 4 1 mol N 2 H g O 2 x 1 mol O 2 x 2 mol N 2 = mol N g O 2 1 mol O 2 N 2 H 4 (l) + O 2 (g) N 2 (g) + 2H 2 O (g) How many moles of nitrogen are produced when 20.0 g of hydrazine is reacted with 30.0 g of oxygen gas?


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