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OB: practice problems for Stoich, class #2

45.0 grams of aluminum metal dissolves into sufficient hydrochloric acid to completely react. How many grams of hydrogen gas form? Write a balanced chemical reaction, get out your periodic table and your Stoichiometry map. Mind the shark. Show all work. WRITE BIG . This takes 3 steps.

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OB: practice problems for Stoich, class #2

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  1. 45.0 grams of aluminum metal dissolves into sufficient hydrochloric acid to completely react. How many grams of hydrogen gas form? Write a balanced chemical reaction, get out your periodic table and your Stoichiometry map. Mind the shark. Show all work. WRITE BIG. This takes 3 steps OB: practice problems for Stoich, class #2

  2. 2Al(S) + 6HCl(AQ) 2AlCl3(AQ) +3H2(G) 45.0 grams of aluminum metal dissolves into sufficient hydrochloric acid to completely react. How many grams of hydrogen gas form?

  3. 2Al(S) + 6HCl(AQ) 2AlCl3(AQ) +3H2(G) 45.0 grams of aluminum metal dissolves into sufficient hydrochloric acid to completely react. How many grams of hydrogen gas form? 45.0 g Al1 1 mole Al27 g Al X = 1.67 moles Al 2x = 5.01 x = 2.51 moles H2 Al21.67H2 3 x MR 2.51 moles H21 2 g H21 mole H2 X = 5.02 grams H2

  4. 2Al + 6HCl 2AlCl3 + 3H2 Using the same reaction… if you want to use up 23.1 moles of HCl, how many formula units of aluminum chloride form? 2 steps. Never change this to 3:1, It’s not worth it, you’ll possibly make a boo boo, and the math works fine 6:2 HCl6AlCl3 2 MR

  5. 2Al + 6HCl 2AlCl3 + 3H2 Using the same reaction… if you want to use up 23.1 moles of HCl, how many formula units of aluminum chloride form? 2 steps. 6x = 46.2 x = 7.70 moles AlCl3 HCl623.1AlCl3 2 x MR 7.70 moles AlCl31 6.02 x 1023 FU AlCl31 mole AlCl3 X = 7.70 x 6.02 x 1023 FU 46.354 x 1023 FU = 46.4 x 1023 = 4.64 x 1024 FU AlCl3

  6. 371.5 grams of candle wax (C21H44) combust. How many liters of CO2 gas form, assume STP for the gas. 3 steps.

  7. 371.5 grams of candle wax (C21H44) combust. How many liters of CO2 gas form, assume STP for the gas. 3 steps. C21H44 + O2 CO2 + H2O

  8. 371.5 grams of candle wax (C21H44) combust. How many liters of CO2 gas form, assume STP for the gas. 3 steps. C21H44 + 32O2 21CO2 + 22H2O Mole ratio 1 : 32 : 21 : 22 371.5 g wax1 1 mole wax296 g wax X = 1.255 moles wax C21H44C 21 x 12 = 252 gH 44 x 1 = 44 g296 g/mole

  9. 371.5 grams of candle wax (C21H44) combust. How many liters of CO2 gas form, assume STP for the gas. 3 steps. C21H44 + 32O2 21CO2 + 22H2O 371.5 g wax1 1 mole wax296 g wax X = 1.255 moles wax wax 11.255CO2 21 x x = 26.36 moles CO2 MR

  10. 371.5 grams of candle wax (C21H44) combust. How many liters of CO2 gas form, assume STP for the gas. 3 steps. C21H44 + 32O2 21CO2 + 22H2O 371.5 g wax1 1 mole wax296 g wax X = 1.255 moles wax wax 11.255CO2 21 x x = 26.36 moles CO2 MR 22.4 L CO21 mole CO2 26.36 moles CO21 x = 590.5 Liters CO2

  11. Using the same wax combustion reaction, if you consume 23.9 moles of oxygen, how many moles of water form? One step. (sometimes it’s easy) Oxygen wata MR

  12. Using the same wax combustion reaction, if you consume 23.9 moles of oxygen, how many moles of water form? One step. (sometimes it’s easy) Oxygen3223.9 wata 22 x MR 32x = 525.8 x = 16.4 moles wata

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  14. Homework… Due now: 12 Reactions Lab Due Friday: Stoich HW #1 Due Monday: MgO Lab (change to 40 Minutes please) Due Tuesday: Stoich HW #3

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