C9H20+14O2→9CO2+10H2O a. How many molecules of nonane will burn in a tank that has 770 mL of oxygen at STP? b. If the container were heated to 78O C before the combustion, what would the pressure be?

Explanation:

C9H2O + 14O2 → 9CO2 + 10H2O

a. How many molecules of nonane will burn in a tank that has 770 mL of oxygen at STP?

From the reaction;

1 mol of nonane reacts with 14 mol of Oxygen

At STP, 1 mol = 22.4 L

Converting 770 ml (0.77 L) to mol;

1 = 22.4

x = 0.77

x = 0.77 / 22.4 = 0.034375 mol

Back to the equation;

1 mol = 14 mol

x mol = 0.034375 mol

Solving for x;

x = 0.034375 * 1 / 14

x = 0.00245535714 mol

1 mol = 6.022 * 10^23 molecules

0.00245535714 mol = x

Solving for x;

x = 0.00245535714 * 6.022 * 10^23

x = 1.4786 * 10^21 molecules

b. If the container were heated to 78O C before the combustion, what would the pressure be?

Initial Pressure P1 = ?

Initial Temperature T1 = 780 +273 = 1053 K

Final Temperature (STP) T2 = 273 K

Final Pressure (STP) P2 = 1 atm

Using gay lusaac's law;

P1 / T1  =  P2 / T2

Inserting the values;

P1 = T1 * P2 / T2

P1 = 1053 * 1 / 273

P1 = 3.857 atm

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