Acetylene(C2H2) burns in oxygen to form carbon monoxide and water. How many liters of oxygen(at STP) are needed to burn 25.0 L of acetylene (at STP)?

To determine the number of liters of oxygen needed to burn 25.0 L of acetylene, we can use the balanced chemical equation and the molar ratios between acetylene and oxygen.

The balanced chemical equation for the combustion of acetylene is as follows:

2 C2H2 + 5 O2 → 4 CO + 2 H2O

According to the equation, we need 5 moles of oxygen to react with 2 moles of acetylene.

To calculate the number of moles of acetylene in 25.0 L, we can use the ideal gas law:

PV = nRT

Given that the volume (V) is 25.0 L and the temperature (T) is at STP (Standard Temperature and Pressure), we can use the values:
P = 1 atm
R = 0.0821 L·atm/(K·mol)

Using the ideal gas law equation, we can solve for the number of moles (n) of acetylene:

n = PV / RT

n = (1 atm) * (25.0 L) / (0.0821 L·atm/(K·mol) * 273.15 K)

n ≈ 1.093 moles of acetylene

Since the molar ratio between acetylene and oxygen is 2:5, we can set up a ratio between moles of acetylene and moles of oxygen:

2 moles of C2H2 : 5 moles of O2

1.093 moles of C2H2 : x moles of O2

By solving for x, we can find the number of moles of oxygen needed:

x = (1.093 moles of C2H2) * (5 moles of O2 / 2 moles of C2H2)

x ≈ 2.733 moles of O2

Now, we can convert the moles of oxygen into liters using the molar volume at STP, which is 22.4 L/mol:

Volume of O2 = (2.733 moles of O2) * (22.4 L/mol)

Volume of O2 ≈ 61.15 L of O2 (rounded to two decimal places)

Therefore, approximately 61.15 liters of oxygen at STP are needed to burn 25.0 liters of acetylene.

To determine the amount of oxygen needed to burn acetylene, we can use the balanced chemical equation for the combustion of acetylene:

C2H2 + 5O2 -> 2CO + 2H2O

From the equation, we can see that 1 mole of acetylene reacts with 5 moles of oxygen to produce 2 moles of carbon monoxide and 2 moles of water.

First, let's find the number of moles of acetylene in 25.0 L:

Using the ideal gas law at STP (Standard Temperature and Pressure), which is 1 atm and 273 K, we can say that 1 mole of any ideal gas occupies 22.4 L.

25.0 L of acetylene at STP is equal to 25.0 L / 22.4 L/mol = 1.12 mol.

Since 1 mole of acetylene requires 5 moles of oxygen, we can calculate the number of moles of oxygen needed:

1.12 mol acetylene × (5 mol oxygen/1 mol acetylene) = 5.6 mol oxygen.

Now, to convert moles of oxygen to liters, we use the ideal gas law again:

5.6 mol oxygen × 22.4 L/mol = 125.44 L oxygen.

Therefore, 25.0 L of acetylene at STP would require 125.44 L of oxygen at STP to undergo complete combustion.

You can take the shortcut when everything is in L and use L as mols.

2C2H2 + 5O2 ==> 4CO2 + 2H2O

L oxygen needed = L C2H2 x (5 mols O2/2 mols C2H2) = 25L x 5/2 = ?