You have .5g of copper. If the nitric acid is 16 M HNO3, how many milliliters of nitric acid will react exactly with that amount of copper. I know that when I get that volume I will triple it so that there will be excess acid and the copper will be the limiting reactant.
.5g Cu is .000787 moles Cu
16M HNO3 is 16 moles/L
You need 2 moles of HNO3 for each mole of Cu, so that's .00157 moles HNO3
.00157 mole / 16mole/L = .0000981 L = .0981mL
Seems like an awfully small volume. Better check my math.
I think 0.5/63.54 = 0.00787 mols Cu.
To determine the volume of 16 M HNO3 that will react with 0.5g of copper, we need to use stoichiometry and the balanced equation for the reaction between copper and nitric acid.
The balanced equation for the reaction is as follows:
3Cu + 8HNO3 -> 3Cu(NO3)2 + 2NO + 4H2O
From the balanced equation, we can see that three moles of copper are required for every eight moles of nitric acid.
First, let's calculate the number of moles of copper we have:
Mass of copper = 0.5g
Molar mass of copper (Cu) = 63.55 g/mol
Number of moles of copper = Mass / Molar mass
Number of moles of copper = 0.5g / 63.55 g/mol = 0.00787 mol (rounded to 4 decimal places)
Based on the stoichiometry, the ratio of moles of copper to moles of nitric acid is 3:8.
Now, we can calculate the number of moles of nitric acid required:
Moles of nitric acid = (Number of moles of copper / 3) * 8
Moles of nitric acid = (0.00787 mol / 3) * 8 = 0.0210 mol (rounded to 4 decimal places)
Finally, we can convert the moles of nitric acid to milliliters using the molarity of the acid.
Molarity (M) = Moles / Volume (L)
We rearrange the equation to isolate volume:
Volume (L) = Moles / Molarity
Let's convert the volume to milliliters:
Volume (mL) = Volume (L) * 1000
Volume (L) = 0.0210 mol / 16 M = 0.0013125 L (rounded to 4 decimal places)
Volume (mL) = 0.0013125 L * 1000 = 1.31 mL (rounded to 2 decimal places)
So, 1.31 mL of 16 M HNO3 will react exactly with 0.5g of copper. To create an excess of acid, you can triple the volume to approximately 3.93 mL.
To determine the volume of nitric acid needed to react with 0.5g of copper, we can use stoichiometry. The balanced equation for the reaction between copper (Cu) and nitric acid (HNO3) is as follows:
3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O
From the equation, we can see that 3 moles of copper react with 8 moles of nitric acid.
1. Convert the mass of copper to moles. The molar mass of copper is approximately 63.55 g/mol:
moles of Cu = mass / molar mass
moles of Cu = 0.5g / 63.55 g/mol
moles of Cu ≈ 0.008 moles
2. Use the mole ratio from the balanced equation to find the moles of nitric acid needed:
moles of HNO3 = moles of Cu × (8 moles of HNO3 / 3 moles of Cu)
moles of HNO3 ≈ 0.008 moles × (8/3)
moles of HNO3 ≈ 0.0213 moles
3. Finally, use the molarity to find the volume of nitric acid in liters:
volume of HNO3 (in L) = moles of HNO3 / molarity
volume of HNO3 (in L) = 0.0213 moles / 16 M
volume of HNO3 (in L) ≈ 0.00133 L
To convert the volume in liters to milliliters, simply multiply by 1000:
volume of HNO3 (in mL) ≈ 0.00133 L × 1000
volume of HNO3 (in mL) ≈ 1.33 mL
Therefore, approximately 1.33 milliliters of 16 M HNO3 will react exactly with 0.5 grams of copper.