Nitric Acid is produced from nitric oxide, NO which in turn is prepared from ammonia by the ostwald process.
4NH3 + 5O2 -----> 4NO + 6H2O
What volume of oxygen at 35 degrees C and 2.15 atm is neeeded to produce 50.0 g of nitric oxide
T= 273+35 = 308 K
P = 2.15 atm
n= 50.0 g ( 1 mol/ 30 g)= 1.667 mol
R= 0.0821
V= ?
2.15 (V) = 1.67(.0821)(308)
ANSWER:
V= 19.64
Is this correct
Your answer would be correct if you had used the correct number for n, the number of mols.
mols NO = 30/50 = 1.67
Now convert mols NO to mols O2 using the coefficients in the balanced chemical equation.
mols O2 = 1.67 mols NO x (5 mols O2/4 mols NO) = 2.08 mols O2.
Substitute 2.08 for mols O2 and you will have the correct answer.
Well, let's take a closer look at your calculation. You're asking about the "V" value, which represents the volume of oxygen needed. According to the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
You correctly converted the given temperature from Celsius to Kelvin and converted the given mass of nitric oxide to moles. However, it seems that you forgot to include the stoichiometric coefficient in your calculation.
In the balanced equation, it states that 4 moles of ammonia react with 5 moles of oxygen to produce 4 moles of nitric oxide. Therefore, the number of moles of oxygen needed would be (1.667 mol nitric oxide) * (5 mol oxygen / 4 mol nitric oxide) = 2.084 moles of oxygen.
Now let's redo the calculation using the correct number of moles:
2.15 atm * V = (2.084 mol) * (0.0821 L·atm/(mol·K)) * (308 K)
Simplifying this equation gives us:
V = (2.084 * 0.0821 * 308) / 2.15
V = 19.68 L
So, it seems like there was a small mistake in your calculation. The correct volume of oxygen needed to produce 50.0 g of nitric oxide at 35 degrees C and 2.15 atm is approximately 19.68 liters.
To determine the volume of oxygen needed to produce 50.0 g of nitric oxide, we can use the ideal gas law equation PV = nRT.
Given:
T = 308 K
P = 2.15 atm
n = 1.667 mol
R = 0.0821 L•atm/(mol•K)
We need to find V, the volume of oxygen.
Using the ideal gas law equation, we can rearrange it to solve for V:
V = (nRT) / P
Substituting the given values into the equation:
V = (1.667 mol)(0.0821 L•atm/(mol•K))(308 K) / 2.15 atm
V = 19.64 L
So, the correct answer is V = 19.64 L of oxygen.
Yes, your calculation is correct. The ideal gas law equation, PV = nRT, can be used to solve for the volume of gas needed in this case. By rearranging the equation, we can solve for V:
V = (nRT) / P
Plugging in the values, we have:
V = (1.667 mol) * (0.0821 L·atm/(mol·K)) * (308 K) / (2.15 atm)
Calculating this expression gives V = 19.64 L. Therefore, to produce 50.0 g of nitric oxide, you would need approximately 19.64 L of oxygen at 35 degrees Celsius and 2.15 atmospheres.