What volume of oxygen gas at STP is needed to react with 5.06 mol of C2H4, according to the following equation C2H4 + 3O2 ----> 2CO2 + 2H2O

5.06 mol C2H4 requires 3*5.06 mols O2. 1 mol O2 at STP occupies 22.4 L; therefore, 3*5.06 mols will occupy ..... L.

340 L

To determine the volume of oxygen gas needed to react with 5.06 mol of C2H4, we can use the stoichiometric coefficients from the balanced equation.

From the balanced equation: C2H4 + 3O2 -> 2CO2 + 2H2O

We can see that the mole ratio between C2H4 and O2 is 1:3. That means for every 1 mol of C2H4, we need 3 moles of O2.

Given that we have 5.06 mol of C2H4, we can multiply this by the mole ratio to find the corresponding amount of O2 needed.

5.06 mol C2H4 * (3 mol O2 / 1 mol C2H4) = 15.18 mol O2

Now, to find the volume of O2 at STP, we can use the molar volume of a gas at STP, which is 22.4 L/mol.

15.18 mol O2 * 22.4 L/mol = 339.07 L

Therefore, the volume of oxygen gas at STP needed to react with 5.06 mol of C2H4 is approximately 339.07 L.

To find the volume of oxygen gas at STP needed to react with 5.06 mol of C2H4, we can use the ideal gas law equation: PV = nRT.

First, we need to calculate the number of moles of oxygen gas required for the reaction. According to the balanced equation, the stoichiometric ratio between C2H4 and O2 is 1:3. Therefore, 1 mol of C2H4 reacts with 3 mol of O2.

Since we have 5.06 mol of C2H4, we can multiply it by the stoichiometric ratio of O2 to determine the number of moles of O2:
5.06 mol C2H4 * (3 mol O2 / 1 mol C2H4) = 15.18 mol O2

Now that we have the number of moles of O2, we can use the ideal gas law to find the volume of O2 at STP. At STP (Standard Temperature and Pressure), the pressure (P) is 1 atm, the volume (V) can be measured in liters, the number of moles (n) is 15.18 mol, and the gas constant (R) is 0.0821 L·atm/(mol·K). The temperature (T) is 273 K.

Plugging the values into the equation:
PV = nRT

V * 1 atm = 15.18 mol * 0.0821 L·atm/(mol·K) * 273 K

V = (15.18 mol * 0.0821 L·atm/(mol·K) * 273 K) / 1 atm

Simplifying:
V = 336.88 L

Therefore, the volume of oxygen gas at STP needed to react with 5.06 mol of C2H4 is 336.88 liters.