To determine whether you have enough gas for the reaction, we can use the ideal gas law:
\[ PV = nRT \]
where:
- \( P \) is the pressure (in kPa)
- \( V \) is the volume (in L)
- \( n \) is the number of moles of gas
- \( R \) is the gas constant (8.31 L-kPa/mol-K)
- \( T \) is the temperature (in K)
First, let's find the number of moles (\( n \)) of the gas you have in the 35.0 L at 100 kPa and 323 K.
Given:
- \( P = 100 , \text{kPa} \)
- \( V = 35.0 , \text{L} \)
- \( T = 323 , \text{K} \)
- \( R = 8.31 , \text{L-kPa/mol-K} \)
Now, we can rearrange the ideal gas law to solve for \( n \):
\[ n = \frac{PV}{RT} \]
Substituting the values:
\[ n = \frac{100 , \text{kPa} \times 35.0 , \text{L}}{8.31 , \text{L-kPa/mol-K} \times 323 , \text{K}} \]
Calculating the denominator:
\[ n = \frac{3500 , \text{kPa·L}}{2677.93 , \text{L-kPa/mol}} \]
\[ n \approx 1.31 , \text{mol} \]
Next, we need to find out how many moles of gas are required to achieve 22.4 L at STP (Standard Temperature and Pressure, which is 0°C or 273 K and 1 atm or approximately 101.3 kPa).
1 mole of an ideal gas occupies 22.4 L at STP, meaning for 22.4 L:
\[ n_{required} = \frac{22.4 , \text{L}}{22.4 , \text{L/mol}} = 1 , \text{mol} \]
Now, comparing the moles you have and the moles required:
- You have \( 1.31 , \text{mol} \)
- You need \( 1 , \text{mol} \)
Since you have more moles than required, you will have an excess of gas, and the reaction will occur.
Therefore, the correct statement is: You will have an excess of gas and the reaction will occur.