Question

When acetylene gas (C2H2) burns, how many liters of O2 at STP are used for every 3.35 mol of acetylene burned? The chemical reaction is 2C2H2 + 5O2 → 4CO2 + 2H2O.

188 L

30.0 L

375 L

37.5 L

Answer 1

188 L.

Step-by-step explanation:

Let's use the balanced chemical equation to determine the stoichiometry of the reaction between acetylene (C2H2) and oxygen (O2):

2C2H2 + 5O2 → 4CO2 + 2H2O

According to the balanced equation, 2 moles of acetylene react with 5 moles of oxygen. This ratio can be used to calculate the amount of oxygen required for a given amount of acetylene.

Given that you have 3.35 moles of acetylene, you can set up a proportion to find the moles of oxygen required:

(3.35 mol C2H2 / 2 mol C2H2) = (x mol O2 / 5 mol O2)

Solving for x (moles of O2):

x = (3.35 mol C2H2 * 5 mol O2) / 2 mol C2H2

x = 8.375 mol O2

Now that we know the moles of oxygen required, we can use the ideal gas law to calculate the volume of oxygen at STP (Standard Temperature and Pressure) that corresponds to 8.375 moles:

V = nRT/P

Where:

- V is the volume of the gas (in liters)

- n is the number of moles of the gas

- R is the ideal gas constant (approximately 0.0821 L.atm/mol.K)

- T is the temperature in Kelvin (STP is 273.15 K)

- P is the pressure in atmospheres (STP is 1 atm)

Plugging in the values:

V = (8.375 mol * 0.0821 L.atm/mol.K * 273.15 K) / 1 atm

V ≈ 189.8 L

Rounded to the nearest whole number, the volume of oxygen required at STP for the given amount of acetylene is approximately 190 L, which is closest to the provided value of 188 L. So, the correct answer is **188 L**.