A chemist mixes 96.8 g of ammonia, NH3, with 188 L of O2 at STP. How many grams of NO will be produced? The balanced equation for this reaction is:
4NH3 +5O2 = 4NO + 6H2O
you will have molesNH3=96.8/17=5.69
according to the balanced equation, you need 5/4 (5.69)moles=7.11moles O2.
you have 188/22.4 molesO2=8.39molesO2, so you have plenty of oxygen. The reaction will exhaust the ammonia.
You should then get 5.69 moles of NO then. Convert that to grams.
To solve this problem, we need to use the stoichiometry of the balanced equation. The coefficients in the balanced equation give us the mole ratio between reactants and products. We can use this mole ratio to calculate the number of moles of nitrogen monoxide (NO) produced, and then convert it to grams using its molar mass.
First, let's calculate the number of moles of ammonia (NH3) and oxygen (O2) using their respective masses and molar masses.
Molar mass of NH3 = 17.03 g/mol
Number of moles of NH3 = mass of NH3 / molar mass of NH3 = 96.8 g / 17.03 g/mol = 5.68 mol
Since the reaction is at STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 L at 0 degrees Celsius and 1 atm. Therefore, we can calculate the number of moles of oxygen using the ideal gas law.
Number of moles of O2 = volume of O2 (in L) / 22.4 L/mol = 188 L / 22.4 L/mol = 8.39 mol
Now, let's use the balanced equation to find the mole ratio between NH3 and NO. According to the equation, 4 moles of NH3 react with 4 moles of NO.
Next, we need to determine the limiting reactant. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed. To do this, we compare the mole ratio of NH3 to NO with the actual mole ratio in the reaction.
Mole ratio of NH3 to NO = 4:4 = 1:1
Actual mole ratio of NH3 to NO = 5.68 mol NH3 / 4 mol NO = 1.42
Since the actual mole ratio is higher than the ratio in the balanced equation, NH3 is in excess and oxygen is the limiting reactant.
The stoichiometry tells us that 5 moles of O2 react to produce 4 moles of NO.
Now, we can calculate the number of moles of NO that will be produced using the mole ratio from the balanced equation.
Number of moles of NO = Number of moles of O2 × (4 mol NO / 5 mol O2) = 8.39 mol × (4/5) = 6.71 mol
Finally, we can convert the moles of NO to grams using its molar mass.
Molar mass of NO = 30.01 g/mol
Mass of NO = Number of moles of NO × molar mass of NO = 6.71 mol × 30.01 g/mol = 201.2 g
Therefore, approximately 201.2 grams of NO will be produced.