Question

Який об'єм карбон (IV) оксиду (н. у.) можна отримати внаслідок спалювання етану об'ємом 35 л?

Answer 1

1,5625. x

2C2H6+7O2-->4CO2+6H2O

n(C2H6)=35:22,4=1,5625

x=1,5625×2=3,125

V=3,125×22,4=70л

Answer 2

Volume of Carbon (IV) Oxide Produced from Burning 35 L of Ethane

To calculate the volume of carbon (IV) oxide (CO2) produced from burning 35 L of ethane, we can use the balanced chemical equation for the combustion of ethane.

The balanced chemical equation for the combustion of ethane is: C2H6 + 7/2 O2 → 2 CO2 + 3 H2O

From the balanced equation, we can see that 1 mole of ethane produces 2 moles of carbon dioxide.

Now, let's calculate the moles of ethane in 35 L using the ideal gas law: PV = nRT Where: P = pressure (atm) V = volume (L) n = moles R = ideal gas constant (0.0821 L·atm/mol·K) T = temperature (K)

Assuming standard temperature and pressure (STP), which is 0°C (273 K) and 1 atm, we can calculate the moles of ethane: n = PV/RT = (1 atm) * (35 L) / (0.0821 L·atm/mol·K * 273 K)

After finding the moles of ethane, we can use the stoichiometry of the balanced chemical equation to find the moles of carbon dioxide produced. Finally, we can convert the moles of carbon dioxide to volume using the ideal gas law.

Let's calculate the volume of carbon (IV) oxide produced from burning 35 L of ethane.

Note: The search results provided do not contain relevant information for this specific calculation. Therefore, the above explanation is based on the principles of stoichiometry and the ideal gas law.