Q2. For the reaction given, how many litres of carbon dioxide gas will be formed by the reaction of 400
litres of oxygen with sufficient pentene (CH..)? The reaction takes place at 125°C and 10.5 atm.
2 C5H10(1) + 15 O2(g) →10 CO2(g) + 10 H20 (8)

266 liters Carbon iv oxide will be formed

Explanation:

The first thing we need to do here is to calculate the number of moles of oxygen reacted.

This can be calculated using the ideal gas equation;

Mathematically;

PV = nRT

From the question;

P = pressure = 10.5 atm

V = volume = 400 liters

n = number of moles = ?

T = temperature = 125 + 273 = 398K

R = molar gas constant = 0.0821 L•atm•K^-1•mol^-1

Substituting these values, we have ;

n = PV/RT

n = (10.25 * 400)/(398 * 0.0821) = 125 moles

From the question;

15 moles oxygen gave 10 moles CO2

125 moles oxygen will give x moles CO2

x = (125 * 10)/15 = 83 moles

Now, we want to know the volume of CO2, present in 83 moles using the given reaction conditions.

Mathematically;

PV = nRT

V = nRT/P

= (83 * 0.0821 * 398)/10.25 = 265.65 which is approximately 266 Liters

the answer is the first option

explanation:

just googled it m8. not that hard

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