3Cu(s) + 8HNO3(aq) → 2NO(g) + 3Cu(NO3)2(aq) + 4H2O(l)
If 15 g of copper metal was added to an aqueous solution containing 6.0 moles of HNO3, how many moles of NO(g) would be produced, assuming a 59.8 % yield.
YOu need 8/3 moles of nitric acid for each mole of copper
so how many moles of copper?
15g/63.5=.236
so we need 8/3 * .236 of nitric acid, and we have....much more than that, so the reaction is controlled by the mass of Copper.
Now, finally, moles of NO:
we get 2/3 mole of NO for each mole of copper, so with the yield we should get
MolesNO=.59(2/3)(.236)
I dont understand
Well, let's see. We have a chemical equation and we need to find how many moles of NO(g) would be produced. So, first let me convert the mass of copper metal to moles.
Molar mass of Cu = 63.55 g/mol
Moles of Cu = Mass of Cu / Molar mass of Cu
Moles of Cu = 15 g / 63.55 g/mol
Now, let's calculate the moles of Cu(NO3)2 that would be produced:
According to the balanced equation, the ratio between Cu and Cu(NO3)2 is 3:3, which means the moles of Cu(NO3)2 would be the same as the moles of Cu.
Moles of Cu(NO3)2 = 15 g / 63.55 g/mol
Now, since you mentioned a yield of 59.8%, we need to calculate the actual moles of NO(g) produced:
Actual moles of NO(g) = Moles of Cu(NO3)2 * (2 Moles of NO(g) / 3 Moles of Cu(NO3)2) * Yield
I hope I didn't lose you with all those numbers and symbols! Let's calculate it:
Actual moles of NO(g) = (15 g / 63.55 g/mol) * (2/3) * 0.598
And now, let's see what we get!
To determine the number of moles of NO(g) produced, we need to use stoichiometry.
First, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and therefore controls the amount of product that can be formed.
Step 1: Calculate the moles of Cu(s) using its molar mass.
Molar mass of Cu = 63.55 g/mol
Moles of Cu = Mass / Molar mass
Moles of Cu = 15 g / 63.55 g/mol ≈ 0.236 mol Cu
Step 2: Calculate the moles of HNO3 using its given quantity.
Moles of HNO3 = 6.0 mol
Step 3: Use the balanced equation to determine the mole ratio between Cu(s) and NO(g).
From the balanced equation: 3 Cu(s) + 8 HNO3(aq) → 2 NO(g) + 3 Cu(NO3)2(aq) + 4 H2O(l)
The mole ratio between Cu(s) and NO(g) is 3:2.
Moles of NO(g) = (Moles of Cu(s) / 3) x 2
Moles of NO(g) = (0.236 mol / 3) x 2 ≈ 0.315 mol NO(g)
Step 4: Calculate the actual yield using the given percentage yield.
Percentage yield = 59.8%
Actual yield = Percentage yield x Theoretical yield
Actual yield = 0.598 x 0.315 mol ≈ 0.188 mol NO(g)
Therefore, approximately 0.188 moles of NO(g) would be produced.