What volume at STP would be required to react with 0.100 mol of hydrogen to produce ammonia?
N2 + 3H2 ==> 2NH3
0.1 mol H2. Look at the coefficients. 3 moles H2 requires 1 mole H2; therefore, 0.1 mol would require.......moles.
Then 1 mole occupies 22.4L at STP so ...mole will occupy .... L.
To determine the volume of a gas at STP (Standard Temperature and Pressure), you'll need to use the ideal gas law. The ideal gas law equation is:
PV = nRT
Where:
P = pressure (in units such as atmospheres)
V = volume (in units such as liters)
n = the number of moles of gas
R = the ideal gas constant (usually 0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
Given that the reaction involves hydrogen gas (H2) and produces ammonia (NH3), we can use stoichiometry to determine the number of moles for this particular reaction.
The balanced equation for the reaction is:
3H2 + N2 → 2NH3
This means that three moles of hydrogen gas react with one mole of nitrogen gas to produce two moles of ammonia.
Since we have 0.100 mol of hydrogen, we can use the stoichiometry to calculate the number of moles of ammonia produced:
0.100 mol H2 × (2 mol NH3 / 3 mol H2) = 0.0667 mol NH3
Now, using the ideal gas law, we can calculate the volume of ammonia at STP.
First, we need to convert the temperature to Kelvin. At STP, the temperature is 273.15 K.
Using the ideal gas law equation:
PV = nRT
Rearranging the equation to solve for V (volume):
V = (nRT) / P
Substituting the values:
V = (0.0667 mol NH3 * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm
Simplifying further:
V = 1.813 L
Therefore, the volume required to react with 0.100 mol of hydrogen to produce ammonia at STP is approximately 1.81 L.