an 18.9g sample of copper and an 82.0 mL of 16.0mol/L nitric acid are allowed to react. Brown nitrogen dioxide gas is generated.
find the maximum mass of NO2 that could be produced.
if 22.6g of the gas is actually produced, find the percentage yield of the reaction.
Cu + 4HNO3 -> Cu(NO3)2 + 2H2O +2NO2
To find the maximum mass of NO2 that could be produced, we need to determine the limiting reactant in the reaction. The limiting reactant is the one that completely reacts and determines the amount of product that can be formed.
First, we need to find the number of moles of copper and nitric acid.
1. Calculate the number of moles of copper (Cu):
- Given mass of copper = 18.9g
- Copper's molar mass (Cu) = 63.55g/mol
- Moles of copper = mass / molar mass = 18.9g / 63.55g/mol
2. Calculate the number of moles of nitric acid (HNO3):
- Given volume of nitric acid = 82.0mL = 0.0820L (convert to liters since the concentration is given in moles per liter)
- Concentration of nitric acid = 16.0mol/L
- Moles of nitric acid = volume x concentration = 0.0820L x 16.0mol/L
3. Use the balanced equation to determine the stoichiometry between copper and nitrogen dioxide.
- According to the equation, 1 mole of copper reacts with 2 moles of nitrogen dioxide.
- So, the mole ratio of copper to nitrogen dioxide is 1:2.
Now, we can determine the limiting reactant:
4. Compare the moles of copper and nitric acid:
- Moles of copper = 18.9g / 63.55g/mol
- Moles of nitric acid = 0.0820L x 16.0mol/L
5. Calculate the moles of nitrogen dioxide that could be formed:
- The mole ratio of copper to nitrogen dioxide is 1:2.
- Moles of nitrogen dioxide = (moles of copper) x 2
6. Determine the limiting reactant:
- The limiting reactant is the one that produces fewer moles of nitrogen dioxide.
- Compare the moles of nitrogen dioxide from copper and nitric acid.
- The reactant that produces fewer moles of nitrogen dioxide is the limiting reactant.
Now that we have the limiting reactant, we can calculate the maximum mass of NO2 that could be produced.
7. Calculate the maximum moles of NO2:
- The moles of nitrogen dioxide are equal to the moles of the limiting reactant.
8. Calculate the maximum mass of NO2:
- Molar mass of NO2 = 46.01g/mol (molar mass of nitrogen + 2 x molar mass of oxygen)
- Mass of NO2 = moles of NO2 x molar mass of NO2
To find the percentage yield of the reaction:
9. Given that 22.6g of NO2 were actually produced, calculate the actual yield in moles:
- Actual yield of NO2 = 22.6g / 46.01g/mol
10. Calculate the percentage yield:
- Percentage yield = (actual yield / theoretical yield) x 100
By following these steps, you can find the maximum mass of NO2 that could be produced and then calculate the percentage yield of the reaction.
mols Cu initially = grams/atomic mass = about 0.3 but you need a more accurate answer.
mols HNO3 initially = M x L = about 1.3 but see note above.
Convert mols Cu to mols NO2. That's about 0.3 mols Cu x (2 mols NO2/1 mols Cu) = about 0.6 mols.
Convert mols HNO3 to mols NO2. That's 1.3 mols HNO3 x (2 mols NO2/4 mols HNO3) = 0.65 (remember is is approx--you go through this more accurately).
In limiting reagent problems the smaller value is ALWAYS the correct value and the reagent producing that number is the limiting reagent. In this Cu is the limiting reagent and HNO3 is the excess reagent.
0.6 mols NO2 produced. Convert to grams. g = mols x molar mass = ? This is the theoretical yield (TY). The actual yield (AY) is 22.5 g.
Then %yield = (AY/TY)*100 = ?
Post your work if you get stuck.