is this correct?

Question

9. What volume of oxygen (in cm3) is required for the complete combustion of 2.32 g of butane?

2 C4H10 (l) + 13 O2 (g)  8 CO2 (g) + 10 H2O (l)

Molar mass of butane = (4 x 12) + (10 x 1) = 58 g/mol

Moles = mass / molar mass = 2.32 / 58 = 0.04 mol

Reacting ratio of C4H10 to O2 = 2 : 13 = 1 : 6.5

Moles of O2 = 0.04 x 6.5 = 0.26 mol

Volume = moles x 24 = 0.26 x 24 = 6.24 cm3

Answer 1

Thought this was a botox question so I will ask the dog as I have no clue

Answer 2

You were doing OK up to the last line where the figure of 24 comes out of nowhere with no units. One mole of an ideal gas will occupy 22.4 Litres at STP. That's 22400 CM3 and 0.26 mol will occupy 5.82 Litres at STP. The question should state at what conditions the volume of O2 is measured and you need to convert using PV = nRT the ideal gas law.

Answer 3

balanced reaction

.. 2 C4H10 + 13 O2 ---> 8 CO2 + 10 H2O 

then

.. 2.32g C4H10... 1 mol C4H10.. .. 13 mol O2

 ----- ---- ---- ---- x ----- ---- ---- ---- x ---- ---- ---- -- = 0.2595 mol O2

.. .... .. .1.. ... .. .. .58.12g C4H10.. .2 mol C4H10

you can't convert that to cm3 of O2 without more information

pressure, temperature, for example

Answer 4

(2.32 g C4H10) / (58.1222 g C4H10/mol) x (13 mol O2 / 2 mol C4H10) =

0.25945 mol O2

You do not specify the conditions under which the volume of O2 is measured.  So supposing 1 atm and 0°C:

(0.25945 mol O2) x (22.414 L/mol) = 5.8153 L = 5.82 L O2

So you are correct up to the last line.  It's not clear to me where "24" actually came from, but it's a reasonable value for liters per mole, using slightly different conditions than I assumed.  But the thing that makes you answer 1000 times wrong, is the unit "cm3".  You have used no units in the calculation, so you are confused about the final answer.  It must be somewhere around 6 LITERS.  You MUST include units in all your calculations.

Answer 5

Butane + Oxygen ➜ Carbon Dioxide + water

2C₄H₁₀ + 13O₂ ➜ 8CO₂ + 10H₂O

molecular weights

C = 12

H = 1

O = 16

2C₄H₁₀ = 2•58 = 116

13O₂ = 13•32 = 416

8CO₂ = 8•44 = 352

10H₂O = 10•18 = 180

check 116+416 = 352+180 = 532

116 grams of C₄H₁₀ + 416 grams of O₂ ➜ 352 grams of CO₂ + 180 grams of H₂O

2 mole of C₄H₁₀ + 13 moles of O₂ ➜ 8 mole of CO₂ + 10 moles of H₂O

gram ratio of oxygen to butane is 416/116

416/116 = x/2.32

x = 8.32 g

but to get volume, you need temperature and pressure.

IF I assume STP, old version,  (but you should so state)

8.32 g / 32 g/mol = 0.26 mol

0.26 mol x 22.41 L/mol = 5.83 L or 5830 mL or 5830 cm³

One mole of any ideal gas at STP has a volume of 22.41L  (old def of STP, 1 atm)STP = 0ºC & 1 ATM   (new 100 kPa)

Answer 6

Yes is correct.