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CH 3 OH + O 2  CO 2 + H 2 O C= H= O= 11 3 4 2 3 4 4 4 21.54232 22 88 88 2 CH 3 OH(g) + 3 O 2 (g)  2 CO 2 (g) + 4 H 2 O(g) O2H2O O2H2O a) Calculate the.

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Presentation on theme: "CH 3 OH + O 2  CO 2 + H 2 O C= H= O= 11 3 4 2 3 4 4 4 21.54232 22 88 88 2 CH 3 OH(g) + 3 O 2 (g)  2 CO 2 (g) + 4 H 2 O(g) O2H2O O2H2O a) Calculate the."— Presentation transcript:

1 CH 3 OH + O 2  CO 2 + H 2 O C= H= O= 11 3 4 2 3 4 4 4 21.54232 22 88 88 2 CH 3 OH(g) + 3 O 2 (g)  2 CO 2 (g) + 4 H 2 O(g) O2H2O O2H2O a) Calculate the number of moles of O 2 consumed if 8 mol of H 2 O is produced. = 3 4 = X 8 4 X = 24 X = 6 mol O 2

2 2 CH 3 OH(g) + 3 O 2 (g)  2 CO 2 (g) + 4 H 2 O(g) b) Calculate the number of grams of O 2 consumed if 8 mol of H 2 O is produced. Mols = mass gfm 6 mol = X 32 g/mol X = 192 grams O 2 2 CH 3 OH(g) + 3 O 2 (g)  2 CO 2 (g) + 4 H 2 O(g) c) Calculate the number of liters of O 2 GAS consumed if 16 liters of H 2 O GAS is produced. O2H2O O2H2O = 3 4 X 16 L = 4 X = 48 X = 12 liter O 2 =

3 2 CH 3 OH(g) + 3 O 2 (g)  2 CO 2 (g) + 4 H 2 O(g) c) Calculate the percentage by mass composition of hydrogen in CH 3 OH. ELEMENTSUBSCRIPT SUMATOMIC MASSSUBTOTAL C 1 mol x12.011 =12.011 H 4 mol x 1.0079 = 4.0316 O 1 mol x15.99 =15.99 + 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 CuSO 4. 5 H 2 O (S) ELEMENTSUBSCRIPT SUMATOMIC MASS (g/MOL) SUBTOTAL Cu 1 mol x63.54 =63.54 H (NON WATER) 0 mol x 1.0079 = 0.0 S 1 mol32.06 H2OH2O 5 mol18.0090.00 O (NON WATER) 4 mol x15.99 =63.69 + 249.2 g/mol % composition = part x 100 whole H 2 O% 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? 22.4 LITERS 44.0 X g 1 LITER = = 1.96 g/L 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 O 2 (g) needed to produce 80.0 L of NO(g)? 4 NH 3 (g) + 5 O 2 (g)  4 NO (g) + 6 H 2 O (g) 80L X O2 O2 NO = = 4 5 4X = 5(80) 4X = 400 X = 100 L NO(g) 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.

7 #28)WHAT IS THE MOLARITY OF A SOLUTION OF KOH WITH 112 GRAMS KOH IN 2.00L OF WATER? MOLARITY = MOLES VOLUME MOLES = MASS GFM MOLES = 112. g = 2mol KNO 3 56 g/mol MOLARITY = 2 mol = 1.00M 2.00L

Download ppt "CH 3 OH + O 2  CO 2 + H 2 O C= H= O= 11 3 4 2 3 4 4 4 21.54232 22 88 88 2 CH 3 OH(g) + 3 O 2 (g)  2 CO 2 (g) + 4 H 2 O(g) O2H2O O2H2O a) Calculate the."

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