Jump to University, Newark -CHEM 1210-Spring18-DOGAN-EKICI Activities and Due Da

Question

Jump to University, Newark -CHEM 1210-Spring18-DOGAN-EKICI Activities and Due Dates HW 12 4/15/2018 11:55 PM 81.9/1004/14/2018 06:10 PM Gradebook O Assignment Inform Available From: No Due Date Points Possible: 10 Grade Category. Gr Sapling Learning The picture below shows two bulbs connected by a stopcock. The large bulb, with a volume of 6.00 L contains nitric axide at a pressuro of 0.550 atm, and the smal bub with a volume of 1.50L, contains oxygen at a pressure of 2.50 atm. The temperature at the beginning and the end of the experiment is 22 c Ater the stopcock is opened, the gases mix and reacr You can check your an You can view solutione You can keep trying t you get it right or give You lose 5% of the po- n your question for e No Which gases are present at the end of the experiment? NO O eTetbook O Help With This Topic O Web Help&videos; O Technical Suppoirt ar 02 NO Whatl are the partial pressures of the gases? If the gas was consumed completely, put O for the answer atm 5 6 3 4

Explanation / Answer

Moles of NO in large bulb = PV/RT

=( 0.550 atm x 6 L) / (0.0821 L-atm/mol-K x 295K)

= 0.136 mol

Moles of O2 in large bulb = PV/RT

=( 2.5 atm x 1.5 L) / (0.0821 L-atm/mol-K x 295K)

= 0.155 mol

Balanced equation

2NO + O2 = 2NO2

2 moles of NO reacts with = 1 mol O2

0.136 moles of NO reacts with = 1*0.136/2 = 0.068 mol O2

Limiting reactant = NO

Excess reactant = O2

At the end of the experiment

O2 and NO2 gases are present

All NO consumed in the reaction

PNO = 0 atm

Moles of O2 at the end = 0.155 - 0.068 = 0.087 mol

Total volume V = 6 + 1.5 = 7.5 L

PO2 = nRT/V

= 0.087 x 0.0821 x 295 / 7.5

= 0.28 atm

Moles of NO2 formed = 0.136 mol

PNO2 = nRT/V

= 0.136 x 0.0821 x 295 / 7.5

= 0.44 atm

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