When 34.5 L ammonia and 39.5 L oxygen gas at STP burn, nitrogen monoxide and water are produced. After the products return to STP, how many grams of nitrogen monoxide are present?
4NH3+5O2->4NO+6H2O
Lets check to see if there is a limiting reageant.
Four volumes of ammonia, five volumes of O2
5/4 *34.5=43 L O2, so there is not enough oxygen.
4/5*39.5=31.6 L of ammonia are reacted.
Liters of NO produced: 31.6L NO
moles: 31.6/22.4
You figure the grams.
To determine the amount of nitrogen monoxide (NO) formed, we can use the balanced chemical equation you provided:
4NH3 + 5O2 -> 4NO + 6H2O
First, let's find the limiting reactant, which is the reactant that is completely consumed and determines the amount of product formed.
To do this, we need to calculate the number of moles for each reactant. We can use the ideal gas law:
PV = nRT
where:
P = pressure (STP = 1 atm)
V = volume (given as 34.5 L for ammonia and 39.5 L for oxygen gas)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (STP = 273 K)
For ammonia (NH3):
n(NH3) = (P * V) / (R * T)
= (1 atm * 34.5 L) / (0.0821 L·atm/(mol·K) * 273 K)
≈ 1.625 moles
For oxygen gas (O2):
n(O2) = (P * V) / (R * T)
= (1 atm * 39.5 L) / (0.0821 L·atm/(mol·K) * 273 K)
≈ 1.895 moles
According to the balanced equation, the ratio of NH3 to NO is 4:4, and the ratio of O2 to NO is 5:4. We can compare the moles of NH3 with the moles of O2 to see which reactant is in excess:
Moles NH3 / Coefficient NH3 = Moles O2 / Coefficient O2
1.625 / 4 = 1.895 / 5
0.40625 ≈ 0.379
Since the ratio is smaller for NH3, it is the limiting reactant.
Next, we can calculate the moles of NO produced. The ratio of NH3 to NO is 4:4, so the moles of NO will be the same as the moles of NH3 used:
Moles NO = 1.625 moles
Now, we can convert the moles of NO to grams using its molar mass. The molar mass of NO is:
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
Mass NO = Moles NO * Molar mass NO
= 1.625 moles * 30.01 g/mol
= 48.77 g
Therefore, after the products return to STP, approximately 48.77 grams of nitrogen monoxide (NO) are present.
To determine the number of grams of nitrogen monoxide (NO) produced, we need to follow these steps:
Step 1: Convert liters of ammonia (NH3) and oxygen gas (O2) to moles.
Given:
Volume of NH3 = 34.5 L
Volume of O2 = 39.5 L
At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore, we can calculate the number of moles of NH3 and O2 as follows:
Moles of NH3 = 34.5 L / 22.4 L/mol
Moles of O2 = 39.5 L / 22.4 L/mol
Step 2: Determine the limiting reactant.
The balanced equation provided indicates that 4 moles of NH3 react with 5 moles of O2 to produce 4 moles of NO. Therefore, we need to compare the moles of NH3 and O2 to identify the limiting reactant.
Using the mole ratios from the balanced equation, we can calculate the moles of NO formed:
Moles of NO = Moles of NH3 (multiplied by mole ratio)
Step 3: Convert moles of NO to grams using the molar mass.
The molar mass of NO is the sum of the atomic masses of nitrogen (N) and oxygen (O) in one mole of NO.
Molar mass of NO = atomic mass of N + 1/2 * atomic mass of O
Using the molar mass, we can convert moles of NO to grams:
Grams of NO = Moles of NO * Molar mass of NO
Following these steps, you can determine the number of grams of nitrogen monoxide produced when 34.5 L of ammonia and 39.5 L of oxygen gas at STP burn.