Copper reacts with nitric acid according to the following reaction:
3 Cu (s) + 8 HNO3 (aq) --> 3 Cu(NO3)2 (aq) + 2 NO (g) + 4 H2O (l)
If an pre 1982 copper penny contains 3.10 grams of copper, what volume of 8.00 M nitric acid is required to exactly consume it? What volume of nitrogen monoxide gas measured at STP would be produced? (Remember that 1 mole of any gas measured at STP = 22.4 dm3)
Well, it seems like we have a chemistry problem on our hands. I may not be the most reliable source for calculations, but let's give it a shot!
To find the volume of 8.00 M nitric acid required to consume 3.10 grams of copper, we need to determine how many moles of copper are present. We can do this by dividing the mass of copper by its molar mass.
The molar mass of copper is approximately 63.55 g/mol, so:
Number of moles of copper = 3.10 g / 63.55 g/mol
Now, we can use the balanced equation to determine the mole ratio between copper and nitric acid. From the equation, we know that 3 moles of copper react with 8 moles of nitric acid.
Therefore, the number of moles of nitric acid required would be:
Number of moles of nitric acid = (Number of moles of copper) / 3 * 8
Now that we have the number of moles of nitric acid, we can use its molarity to find the volume. Remember that Molarity (M) is moles of solute per liter of solution.
Volume of 8.00 M nitric acid = (Number of moles of nitric acid) / 8.00 M
As for the volume of nitrogen monoxide gas (NO) produced, we can use the mole ratio between copper and NO from the balanced equation, which is 3 moles of copper to 2 moles of NO.
First, we find the number of moles of NO produced:
Number of moles of NO = (Number of moles of copper) / 3 * 2
Finally, we can convert the moles of NO to volume using the molar volume at STP:
Volume of NO gas = (Number of moles of NO) * 22.4 dm^3/mol
Please note that these calculations assume complete reaction and ideal conditions. It's always a good idea to double-check the numbers and consult an expert if needed.
I hope my clownish attempt at answering this chemistry question brings a smile to your face!
To determine the volume of 8.00 M nitric acid required to completely consume a pre-1982 copper penny, we first need to calculate the number of moles of copper in the penny.
Given:
Mass of copper = 3.10 grams
The molar mass of copper (Cu) is 63.546 g/mol. Therefore, the number of moles of copper can be calculated as follows:
Number of moles of copper = mass of copper / molar mass of copper
= 3.10 g / 63.546 g/mol
Now we can determine the molar ratio between copper and nitric acid. From the balanced chemical equation, we know that 3 moles of copper react with 8 moles of nitric acid.
From the molar ratio, we can calculate the number of moles of nitric acid required:
Number of moles of nitric acid = (Number of moles of copper) x (8 moles of nitric acid / 3 moles of copper)
Now we can calculate the volume of 8.00 M nitric acid required using the following formula:
Volume of nitric acid (in liters) = Number of moles of nitric acid / Molarity of nitric acid
Given that 1 Molar (M) means 1 mol/L, we can use this information to calculate the volume of nitric acid.
Now, to calculate the volume of nitrogen monoxide (NO) gas produced measured at STP:
From the balanced chemical equation, we know that the molar ratio between NO gas and copper is 2 moles of NO gas to 3 moles of copper.
Given that 1 mole of any gas measured at STP is equal to 22.4 dm3:
Volume of NO gas (in liters) = Number of moles of copper x (2 moles of NO gas / 3 moles of copper) x 22.4 dm3/mol
Please substitute the calculated values into the equations above to find the final answers.
To find the volume of 8.00 M nitric acid required to consume 3.10 grams of copper, we need to use stoichiometry. Here are the steps to solve the problem:
Step 1: Convert the mass of copper to moles.
- The molar mass of copper (Cu) is 63.55 g/mol.
- Divide the given mass of copper (3.10 g) by the molar mass to get the number of moles: 3.10 g / 63.55 g/mol = 0.0488 mol.
Step 2: Use the stoichiometry of the balanced chemical equation to determine the mole ratio between copper (Cu) and nitric acid (HNO3).
- From the balanced equation, the mole ratio between Cu and HNO3 is 3:8. This means that 3 moles of Cu react with 8 moles of HNO3.
- Multiply the number of moles of Cu (0.0488 mol) by the mole ratio: 0.0488 mol Cu * (8 mol HNO3 / 3 mol Cu) = 0.131 mol HNO3.
Step 3: Calculate the volume of 8.00 M nitric acid.
- The concentration of the nitric acid (HNO3) is given as 8.00 M. This means there are 8.00 moles of HNO3 in 1 liter of solution.
- Multiply the number of moles of HNO3 (0.131 mol) by the volume concentration: 0.131 mol HNO3 * (1 L / 8.00 mol HNO3) = 0.0164 L.
- Convert the volume from liters to milliliters: 0.0164 L * 1000 = 16.4 mL.
- Therefore, the volume of 8.00 M nitric acid required to consume 3.10 grams of copper is 16.4 mL.
To find the volume of nitrogen monoxide gas (NO) measured at STP produced, we need to use the stoichiometry again:
Step 1: Use the mole ratio between Cu and NO from the balanced equation.
- From the balanced equation, the mole ratio between Cu and NO is 3:2. This means that 3 moles of Cu react to produce 2 moles of NO.
- Multiply the number of moles of Cu (0.0488 mol) by the mole ratio: 0.0488 mol Cu * (2 mol NO / 3 mol Cu) = 0.0325 mol NO.
Step 2: Convert the number of moles of NO to volume at STP.
- According to Avogadro's Law, 1 mole of any gas at STP occupies a volume of 22.4 dm3 (or 22.4 liters).
- Multiply the number of moles of NO (0.0325 mol) by the volume conversion factor: 0.0325 mol NO * (22.4 L / 1 mol NO) = 0.728 L.
- Therefore, the volume of nitrogen monoxide gas produced at STP is 0.728 L or 728 mL.
Convert 3.1 g Cu into moles. moles = grams/molar mass
Using the coefficients in the balanced equation, convert moles Cu to moles HNO3.
Now convert moles HNO3, remembering M = moles/L to L.
From moles Cu, convert to moles NO formed, then moles x 22.4 L/mol = L NO at STP.