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# CH 3 OH + O 2  CO 2 + H 2 O C= C=

CH 3 OH + O 2  CO 2 + H 2 O C= C= H= H= O= O=. 2. 3. 2. 4. 1.5. 2. 2. 2. 8. 8. 8. 8. 1. 1. 2. 4. 4. 3. 4. 4. 3. Download Presentation ## CH 3 OH + O 2  CO 2 + H 2 O C= C=

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1. CH3OH + O2 CO2 + H2O C= C= H= H= O= O= 2 3 2 4 1.5 2 2 2 8 8 8 8 1 1 2 4 4 3 4 4 3 2CH3OH(g) + 3 O2(g)  2 CO2(g) + 4 H2O(g) a) Calculate the number of moles of O2 consumed if 8 mol of H2O is produced. O2 H2O 3 4 X 8 4X = 24 X = 6 mol O2 = =

2. 2CH3OH(g) + 3 O2(g)  2 CO2(g) + 4 H2O(g) b) Calculate the number of grams of O2 consumed if 8 mol of H2O is produced. Mols = mass gfm 6 mol = X 32 g/mol X = 192 grams O2 2CH3OH(g) + 3 O2(g)  2 CO2(g) + 4 H2O(g) c) Calculate the number of liters of O2 GAS consumed if 16 liters of H2O GAS is produced. O2 H2O 3 4 X 16 L 4X = 48 X = 12 liter O2 = = =

3. 2CH3OH(g) + 3 O2(g)  2 CO2(g) + 4 H2O(g) c) Calculate the percentage by mass composition of hydrogen in CH3OH. + 32.0326 g/mol % composition = part x 100 whole H% composition = 4.0316 x 100 = 12.6% 32.0326

4. c) Calculate the percentage by mass composition of WATER in CuSO4.5 H2O (S) + 249.2 g/mol % composition = part x 100 whole H2O% composition = 90.00 x 100 = 36.11% 249.2

5. #16) The GMW (GFW, molar mass) of a gas is 44.0 grams at STP, what is the density? X g 44.0 = 1.96 g/L = 22.4 LITERS 1 LITER DENSITY RATIO, 1 MOLE = 22.4L THE MASS OF 22.4 L IS THE MOLAR MASS AT STP. IN THIS QUESTION THAT IS 44.0 GRAMS (GIVEN).

6. #20)What is the liter volume of O2 (g) needed to produce 80.0 L of NO(g)? 4 NH3 (g) + 5 O2(g)  4 NO (g) + 6 H2O (g) 5 O2 X = = 4 80L NO GAS VOLUMES CAN EXIST IN A RATIO AS DEFINED BY THE COEFFICIENTS OF THE BALANCED EQUATION, RATIO AS YOU WOULD RATIO MOLES! THIS ASSUMES P AND T ARE CONSTANT. 4X = 5(80) 4X = 400 X = 100 L NO(g)

7. #28)WHAT IS THE MOLARITY OF A SOLUTION OF KOH WITH 112 GRAMS KOH IN 2.00L OF WATER? MOLES = MASS GFM MOLARITY = MOLES VOLUME MOLES = 112. g = 2molKNO3 56 g/mol MOLARITY = 2 mol =1.00M 2.00L

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