How many miles of vapor are formed when 10 liters of butane gas, C4H10 is burned in oxygen at STP?

2C4H10 +13H2O=8Co2 +10H2O

10\44.8=X\10
X=10×10\44.8
X=100\44.8
X=2.23

Can't answer with just one dimension of miles. Perhaps you meant "moles"..

2C4H10 + 13O2 ==> 8CO2 + 10H2O
10 L C4H10 will give you 40 L CO2 vapor and 50 L H2O vapor.or a total of 90 L gas @ STP. 1 mol occupies 22.4L so convert 90 L to mols with the factory of 22.4 L.

2C4H10 +13H2O=8Co2 +10H2O

10\44.8=X\10
X=10×10\44.8
X=100\44.8
X=2.23
i.e If the mile is mole

To calculate the number of miles of vapor formed when 10 liters of butane gas, C4H10, is burned in oxygen at STP (Standard Temperature and Pressure), we need to use the balanced chemical equation for the combustion of butane. The balanced equation is as follows:

2 C4H10 + 13 O2 → 8 CO2 + 10 H2O

From the balanced equation, we can see that for every 2 moles of butane burned, we produce 8 moles of carbon dioxide (CO2) and 10 moles of water (H2O).

First, let's calculate the number of moles of butane in 10 liters. To do this, we need to use the ideal gas law:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)

At STP, the temperature is 273 K, and the pressure is 1 atm. The volume is given as 10 liters. Plugging these values into the equation, we can solve for n:

n = (PV) / (RT)
n = (1 atm) * (10 liters) / ((0.0821 L·atm/mol·K) * (273 K))
n ≈ 0.459 moles of butane

Since the balanced equation shows that 2 moles of butane produce 8 moles of CO2 and 10 moles of H2O, we can use this ratio to find the moles of CO2 and H2O produced.

For CO2:
n(CO2) = (0.459 moles butane) * (8 moles CO2 / 2 moles butane)
n(CO2) ≈ 1.836 moles CO2

For H2O:
n(H2O) = (0.459 moles butane) * (10 moles H2O / 2 moles butane)
n(H2O) ≈ 2.295 moles H2O

Next, we convert the moles of CO2 and H2O to miles using their respective molar masses:

Molar mass of CO2 = 44.01 g/mol
Molar mass of H2O = 18.015 g/mol

To convert the moles to grams, we use the formula:

mass = moles * molar mass

For CO2:
mass(CO2) = (1.836 moles CO2) * (44.01 g/mol)
mass(CO2) ≈ 80.724 g CO2

For H2O:
mass(H2O) = (2.295 moles H2O) * (18.015 g/mol)
mass(H2O) ≈ 41.508 g H2O

Finally, we convert grams to miles using the known conversion factor of 1 mile = 1610.3 grams:

For CO2:
miles(CO2) = (80.724 g CO2) / (1610.3 g/mile)
miles(CO2) ≈ 0.0502 miles CO2

For H2O:
miles(H2O) = (41.508 g H2O) / (1610.3 g/mile)
miles(H2O) ≈ 0.0257 miles H2O

Therefore, when 10 liters of butane gas is burned in oxygen at STP, approximately 0.0502 miles of CO2 and 0.0257 miles of H2O vapor are formed.