A reaction yields 3.75 L of nitrogen monoxide. The Volume is measured at 19 *C and a pressure of 1.10 atm. What mass of NO was produced by the reaction?
Plugging those numbers into the equation we get 5.76
To determine the mass of NO produced by the reaction, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in L)
n = amount of substance (in moles)
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
First, we need to convert the temperature from degrees Celsius to Kelvin:
T(°C) + 273.15 = T(K)
19°C + 273.15 = 292.15 K
Next, we can rearrange the ideal gas law equation to solve for the amount of substance (n):
n = PV / RT
n = (1.10 atm * 3.75 L) / (0.0821 L.atm/mol.K * 292.15 K)
n = 14.91 mol
The balanced chemical equation for the reaction would help us determine the stoichiometry. Without it, we cannot directly relate the moles of nitrogen monoxide to the mass of nitrogen monoxide produced.
Once we have the balanced chemical equation, we can use the molar mass of nitrogen monoxide (NO) to convert the moles of NO to grams.
To find the mass of nitrogen monoxide (NO) produced by the reaction, we need to use the Ideal Gas Law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, let's convert the temperature from Celsius to Kelvin. Simply add 273 to the Celsius temperature:
T = 19°C + 273 = 292 K
Next, rearrange the Ideal Gas Law equation to solve for the number of moles (n):
n = PV / RT
From the problem statement, we have:
P = 1.10 atm (pressure)
V = 3.75 L (volume)
T = 292 K (temperature)
R = 0.0821 L•atm/(mol•K) (ideal gas constant)
Now, substitute these values into the equation to find the number of moles of NO:
n = (1.10 atm * 3.75 L) / (0.0821 L•atm/(mol•K) * 292 K)
Calculate this expression to find the number of moles of NO. Let's work it out step by step:
1.10 atm * 3.75 L = 4.125 atm·L
0.0821 L·atm/(mol·K) * 292 K = 23.9572 L·atm/(mol·K)
n = 4.125 atm·L / 23.9572 L·atm/(mol·K)
Now, divide the units to get the moles of NO:
n = 0.172 moles of NO
Finally, calculate the molar mass of NO, which is the sum of the atomic masses:
N = 14.01 g/mol
O = 16.00 g/mol
Molar mass of NO = 14.01 g/mol + 16.00 g/mol = 30.01 g/mol
To find the mass of NO produced, multiply the number of moles by the molar mass:
Mass of NO = 0.172 moles * 30.01 g/mol
Calculate this product to find the mass of NO produced by the reaction.