Hydrogen cyanide gas is produced industrially by the reaction
2 CH4(g) + 3 O2(g) + 2 NH3(g) → 2 HCN(g) + 6 H2O(g)
What volume of O2(g) is needed to completely react 17 L of CH4(g) at STP?
When one has gases one may treat L as if they were moles and not go through the conversion to mols step.
17 L CH4 x (3 mols O2/2 mols CH4) = 17 x 3/2 = ? L O2.
To determine the volume of O2(g) needed to completely react with 17 L of CH4(g) at STP, we can use the stoichiometry of the balanced chemical equation.
According to the balanced chemical equation:
2 CH4(g) + 3 O2(g) + 2 NH3(g) → 2 HCN(g) + 6 H2O(g)
We can see that the stoichiometric ratio between CH4 and O2 is 2:3. This means that for every 2 moles of CH4, we need 3 moles of O2.
First, let's calculate the number of moles of CH4 in 17 L. We can use the ideal gas law equation:
PV = nRT
Assuming STP conditions (standard temperature and pressure):
P = 1 atm
V = 17 L
n = moles of CH4
R = 0.0821 L·atm/mol·K
T = 273 K
Using the ideal gas law equation:
n = PV / RT
= (1 atm) * (17 L) / (0.0821 L·atm/mol·K * 273 K)
≈ 0.727 moles of CH4
Since the stoichiometric ratio of CH4 to O2 is 2:3, we can calculate the number of moles of O2 needed:
moles of O2 = (3/2) * moles of CH4
= (3/2) * 0.727 moles
≈ 1.09 moles of O2
Now, to convert moles of O2 to volume at STP, we can again use the ideal gas law:
n = moles of O2
V = volume of O2
P = 1 atm
R = 0.0821 L·atm/mol·K
T = 273 K
V = nRT / P
= (1.09 moles) * (0.0821 L·atm/mol·K) * (273 K) / (1 atm)
≈ 24.7 L
Therefore, approximately 24.7 L of O2(g) is needed to completely react with 17 L of CH4(g) at STP.
To determine the volume of O2 gas needed to completely react with 17 L of CH4 gas at STP (Standard Temperature and Pressure), we need to use the balanced equation for the reaction provided:
2 CH4(g) + 3 O2(g) + 2 NH3(g) → 2 HCN(g) + 6 H2O(g)
From the balanced equation, we can see that 2 moles of CH4 react with 3 moles of O2. To solve for the volume of O2 needed, we'll follow these steps:
Step 1: Convert liters (volume) of CH4 gas to moles.
Using the relationship between volume and the molar volume at STP (22.4 L/mol), we can calculate the number of moles of CH4 gas.
17 L CH4 * (1 mol CH4 / 22.4 L) = 0.76 mol CH4
Step 2: Determine the moles of O2 needed.
Since the moles of CH4 and O2 are in a ratio of 2:3, we can find the moles of O2 by multiplying the moles of CH4 gas by the appropriate ratio:
0.76 mol CH4 * (3 mol O2 / 2 mol CH4) = 1.14 mol O2
Step 3: Convert moles of O2 to volume of O2 at STP.
Again, using the relationship between volume and the molar volume at STP, we can calculate the volume of O2 gas:
1.14 mol O2 * (22.4 L / 1 mol O2) = 25.46 L O2
Therefore, approximately 25.46 L of O2 gas is needed to completely react with 17 L of CH4 gas at STP.