gases hw days 1 thru 3.notebook - ms. kissinger · 2018-09-05 · gases hw days 1 thru 3.notebook 2...

7
Gases HW days 1 thru 3.notebook 1 March 10, 2014

Upload: others

Post on 08-Apr-2020

2 views

Category:

Documents


0 download

TRANSCRIPT

Page 1: Gases HW days 1 thru 3.notebook - Ms. Kissinger · 2018-09-05 · Gases HW days 1 thru 3.notebook 2 March 10, 2014 d) mass of reactants 1.50 + 2.75 = 4.25 g mass of products 2.06

Gases HW days 1 thru 3.notebook

1

March 10, 2014

Page 2: Gases HW days 1 thru 3.notebook - Ms. Kissinger · 2018-09-05 · Gases HW days 1 thru 3.notebook 2 March 10, 2014 d) mass of reactants 1.50 + 2.75 = 4.25 g mass of products 2.06

Gases HW days 1 thru 3.notebook

2

March 10, 2014

d) mass of reactants 1.50 + 2.75 = 4.25 g

mass of products 2.06 g NO + 1.86g H2O + .33 gNH3 left = 4.25g

2.75 g O2/32gO2 x 6molH2O/5 mol O2 x 18gH2O/molH2O= 1.86 g H2O

Page 3: Gases HW days 1 thru 3.notebook - Ms. Kissinger · 2018-09-05 · Gases HW days 1 thru 3.notebook 2 March 10, 2014 d) mass of reactants 1.50 + 2.75 = 4.25 g mass of products 2.06

Gases HW days 1 thru 3.notebook

3

March 10, 2014

Page 4: Gases HW days 1 thru 3.notebook - Ms. Kissinger · 2018-09-05 · Gases HW days 1 thru 3.notebook 2 March 10, 2014 d) mass of reactants 1.50 + 2.75 = 4.25 g mass of products 2.06

Gases HW days 1 thru 3.notebook

4

March 10, 2014

Page 5: Gases HW days 1 thru 3.notebook - Ms. Kissinger · 2018-09-05 · Gases HW days 1 thru 3.notebook 2 March 10, 2014 d) mass of reactants 1.50 + 2.75 = 4.25 g mass of products 2.06

Gases HW days 1 thru 3.notebook

5

March 10, 2014

111) C, H, N, O.1023 g compound + O 2 ---> .2766g CO 2 = .0991g H 2O.4831 g compd ---> 27.6mL of dry N2 gas at STP.0276L N2 at STP x 1molN2 x 28.0 g N2 = .0345g N2

22.4LN2 1 mol N2

%C = .2766 g CO2 x 12.0 gC = 0.754gC / .1023 g compd 44.0 g CO2 = 73.7%C

%H = 0.0994 g H2O x 2 g H = 0.0110 gH /.1023g compd 18.0 g H2O = 10.75 % H

% N = .0345 g N / .4831 g compd = 7.14 % N

% O = 100.0 - (73.7 + 10.75 + 7.14) = 8.41 % O

73.7 g C x 1 mol C = 6.14/.51 = 12 12 g C10.8 g H x 1 mol H = 10.8/.51 = 211 g H7.14 g N x 1 mol N = .51/.51 = 1C12H21ON14 g N8.41 g O x 1 mol O = .53/.51 = 116 g O

D = MM*P RT

C12H21ON empirical mass =12(12) + 21 + 16 + 14 = 195

392/195 = 2 molecular formula = C24H42O2N2

4.02g/L = MM * (256/760).08206 * 400K

MM = 392 g/mol

Page 6: Gases HW days 1 thru 3.notebook - Ms. Kissinger · 2018-09-05 · Gases HW days 1 thru 3.notebook 2 March 10, 2014 d) mass of reactants 1.50 + 2.75 = 4.25 g mass of products 2.06

Gases HW days 1 thru 3.notebook

6

March 10, 2014

Page 7: Gases HW days 1 thru 3.notebook - Ms. Kissinger · 2018-09-05 · Gases HW days 1 thru 3.notebook 2 March 10, 2014 d) mass of reactants 1.50 + 2.75 = 4.25 g mass of products 2.06

Gases HW days 1 thru 3.notebook

7

March 10, 2014