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Stoichiometry Unit Review Game

Stoichiometry Unit Review Game. Team Name. Come up with a team name It must be related to science It must be appropriate You DO NOT want to give me an inappropriate team name. Rules. Within each round: There are 3 questions per round! You can wager 5, 10, or 15 points

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Stoichiometry Unit Review Game

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  1. Stoichiometry UnitReview Game

  2. Team Name • Come up with a team name • It must be related to science • It must be appropriate • You DO NOT want to give me an inappropriate team name

  3. Rules • Within each round: • There are 3 questions per round! • You can wager 5, 10, or 15 points • You may use the 5, 10 or 15 ONCE per round • Once you use a point value, you may not use it again that round. • You have one song length to turn in your answer • It is ALL or NOTHING!

  4. Question 1 • Calculate the molar mass of the following compounds (INCLUDE UNIT!) • (NH4)2SO4 • 132.14 g/mol • Mn2Se7 • 662.60 g/mol • H3PO4 • 98.00 g/mol

  5. Question 2 Balance the equation. Then calculate the number of grams of iron, Fe, that can be produced is 25.0g of pure Fe2O3 is used. ____Fe2O3+ ___C ___Fe + ___CO 3 1 3 2 1 mol Fe2O3 159.69g Fe2O3 2 mol Fe 1 mol Fe2O3 55.85 g Fe 1 mol Fe 25.0 g Fe2O3 X X X = 17.487 g Fe

  6. Question 3 • Convert the following substances: • How many moles are in 22 grams of argon? • .55 moles Argon • How many grams are in 88.1 moles of magnesium? • 2,140.83 grams Mg • How many moles are in 9.8 grams of calcium? • 0.245 moles Ca

  7. Question 4 Find the theoretical yield of Silicon nitride, Si3N4, if 555g of Silicon is reacted with excess nitrogen. 3Si + 2N2 Si3N4 1 mol Si 28.09g Si 1 mol Si3N4 3 mol Si 140.28 Si3N4 1 mol Si3N4 555g Si X X X = 923.88g

  8. Question 5 • If the theoretical yield is 923.88g of Si3N4 and the actual yield is 775.5 Si3N4. Calculate the percent yield. • 83.94% • Balance the following equation: ___Si + ___N2___Si3N4 3 2 1

  9. Question 6 • C3H8 + 5O2 3CO2 + 4H2O • If I start with 5g of C3H8, what is my theoretical yield of water? • I got a percent yield of 75%. How many grams of water did I make? • X = 6.13 g H2O 1 mol C3H8 44.10 g 4 mol H2O 1 mol C3H8 18.01 g H2O 1 mol H2O 5g C3H8 X X X = 8.17 g

  10. Question 7 • Balance the equation for the reaction of iron (iii) phosphate, FePO4, with sodium sulfate, Na2SO4, to make iron (iii) sulfate, Fe2(SO4)3, and sodium phosphate, Na3PO4. __ FePO4 + ___ Na2SO4__Fe2(SO4)3 + __ Na3PO4 • If I perform this reaction wit 25 grams of iron (iii) phosphate and an excess of sodium sulfate, how many grams of iron (iii) sulfate can I make? • 33.14 g Fe2(SO4)3 • If 18.5 g of iron (iii) sulfate are actually made when I do this reaction, what is my percent yield? • 55.8% 1 2 3 2

  11. Question 7 Answers 1 molFePO4 150.82 g FePO4 399.88 g Fe2(SO4)3 1 molFe2(SO4)3 1 mol Fe2(SO4)3 2 mol FePO4 25g FePO4 X X X = 33.14 g

  12. Question 8 • How many moles of water are produced from the combustion of 12.5 moles of methane, CH4? CH4 + 2O2 CO2 + 2H2O 1 molCH4 16.04 g CH4 18.01 g H2O 1 mol H2O 2 mol H2O 1 molCH4 12.5 g CH4 X X X = 28.07 g

  13. Question 9 • How many grams of hydrogen are required to produce 0.6 moles of methane, CH4? C + 2H2 CH4 2 mol H2 1 mol CH4 2.02 H2 1 mol H2 0.6 molCH4 X X = 2.42 g

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