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To completely react 0.40 mol of magnesium hydroxide, 0.80 mol of hydrochloric acid will be required. Mg(OH) 2 + 2HCl → 2H 2 O + MgCl 2. 0.40 mol Mg(OH) 2 x 2 mol HCl = 0.8 mol HCl 1 mol Mg(OH) 2.
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To completely react 0.40 mol of magnesium hydroxide, 0.80 mol of hydrochloric acid will be required. Mg(OH)2 + 2HCl → 2H2O + MgCl2 0.40 mol Mg(OH)2x 2 mol HCl= 0.8 mol HCl 1 mol Mg(OH)2
How many moles of nitrogen are produced when 20.0 g of hydrazine is reacted with 30.0 g of oxygen gas? N2H4 (l) + O2 (g) → N2 (g) + 2H2O (g) 20.0 g N2H4 x 1 mol N2H4x 1 mol N2= 0.624 mol N2 32.06 g N2H4 1 mol N2H4 30.0 g O2 x 1 mol O2x 1mol N2= 0.938 mol N2 32.00 g O21 mol O2
What mass of silver is produced when 0.500 g of copper shavings are added to a silver nitrate solution containing 1.95 g of silver nitrate? Cu (s) + 2AgNO3(aq) → Cu(NO3)2 (aq) + 2Ag(s) 0.500 g Cu x 1 mol Cux 2 mol Agx 107.90 g Ag 63.55 g Cu 1 mol Cu 1 mol Ag = 1.70 g Ag 1.95 g AgNO3 x 1 mol AgNO3 x 2 mol Agx 107.90 g Ag 169.91 g AgNO32 mol AgNO3 1 mol Ag = 1.24 g Ag
How many grams of the excess reactant are left over when this reaction is complete? Cu (s) + 2AgNO3(aq) → Cu(NO3)2 (aq) + 2Ag(s) 1.24 g Ag x 1 mol Agx 1 mol Cu x 63.55 g Cu = 0.365 g Cu 107.90 g Ag 2 mol Ag 1mol Cu 0.500g Cu - 0.365g Cu = 0.135 g Cu
If a student really performed this reaction in the lab and only collected 1.17g of silver, what would be her percent yield? Cu (s) + 2AgNO3(aq) → Cu(NO3)2 (aq) + 2Ag(s) % Yield = Actual Yield x 100 Theoretical Yield % Yield = 1.17 g Ag x 100 = 94.4% 1.24 g Ag
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 prelabstoichiometric calculations predict that the theoretical yield should be 12.5 g. What's the student's likely actual yield? % Yield = Actual Yield x 100 Theoretical Yield 0.40 = Actual Yield 12.5 g Actual Yield = (0.40) (12.5g) = 5.0 g