Copper metal reacts with concentrated nitric acid according to the following balanced equation:
3Cu(s) + 8HNO3(aq) �¨ 3Cu(NO3)2(s) + 2NO(g) + 4H2O(l)
Calculate the mass in grams of the excess reagent remaining after the complete reaction of 4.63 g of Cu with 9.26 g of HNO3. Round your answer to 2 decimal places.
1. You have the balanced equation.
2. Convert g Cu to mols Cu.
3. Convert g HNO3 to mols HNO3.
4a. Convert mols Cu to mols Cu(NO3)2.
4b. Convert mols HNO3 to mols Cu(NO3)2.
4c. Choose the smaller mols of Cu(NO2)2 2. The reagent producing that number (either 4a or 4b) will be the limiting reagent.
5. Convert mols of the limiting reagent to mols of the other reagent. That will be how much of the other reagent (the excess reagent) is used.
6. Convert mols of the excess reagent used to grams, subtract from the initial mass present, and that will give you the amount remaining after reaction.
Post your work if you need more help.
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To calculate the mass of the excess reagent remaining after the complete reaction, we need to determine the limiting reagent first.
Step 1: Calculate the moles of Cu and HNO3
Molar mass of Cu = 63.55 g/mol
Molar mass of HNO3 = 63.01 g/mol
moles of Cu = mass of Cu / molar mass of Cu
= 4.63 g / 63.55 g/mol
= 0.073 mol
moles of HNO3 = mass of HNO3 / molar mass of HNO3
= 9.26 g / 63.01 g/mol
= 0.147 mol
Step 2: Determine the limiting reagent
According to the balanced equation, the molar ratio between Cu and HNO3 is 3:8.
Therefore, the moles of Cu and HNO3 have a ratio of 1:8/3.
moles of HNO3 = (8/3) * moles of Cu
0.147 mol = (8/3) * 0.073 mol
Since the actual moles of HNO3 is greater than the moles required, Cu is the limiting reagent.
Step 3: Calculate the moles of Cu(NO3)2 produced
From the balanced equation, it is established that 3 moles of Cu reacts to produce 3 moles Cu(NO3)2.
moles of Cu(NO3)2 = 3 * moles of Cu
= 3 * 0.073 mol
= 0.219 mol
Step 4: Calculate the mass of Cu(NO3)2 produced
Molar mass of Cu(NO3)2 = 63.55 g/mol + 2 * (14.01 g/mol + 16.00 g/mol * 3)
= 187.57 g/mol
mass of Cu(NO3)2 = moles of Cu(NO3)2 * molar mass of Cu(NO3)2
= 0.219 mol * 187.57 g/mol
= 41.02 g
Step 5: Calculate the excess reagent
mass of excess HNO3 = initial mass of HNO3 - mass of HNO3 used
= 9.26 g - 0.147 mol * (63.01 g/mol)
= 9.26 g - 9.252 g
= 0.01 g
Therefore, the mass of the excess reagent remaining after the complete reaction is 0.01 grams.
To solve this problem, we need to determine the limiting reagent and calculate the mass of the excess reagent remaining.
1. Start by calculating the moles of Cu and HNO3 using their respective masses and molar masses:
Moles of Cu = Mass of Cu / Molar mass of Cu
Moles of HNO3 = Mass of HNO3 / Molar mass of HNO3
Given:
Mass of Cu = 4.63 g
Molar mass of Cu = 63.55 g/mol
Mass of HNO3 = 9.26 g
Molar mass of HNO3 = 63.01 g/mol
Moles of Cu = 4.63 g / 63.55 g/mol
= 0.073 mol
Moles of HNO3 = 9.26 g / 63.01 g/mol
= 0.147 mol
2. Next, use the balanced equation to determine the mole ratio between Cu and HNO3. From the equation:
3 moles of Cu react with 8 moles of HNO3.
Calculate the expected moles of HNO3 needed to react with the moles of Cu:
Moles of HNO3 needed = (3/8) * Moles of Cu
= (3/8) * 0.073 mol
= 0.027375 mol
3. Compare the moles of HNO3 needed (step 2) with the actual moles of HNO3 used (step 1). The lower value will be the limiting reagent.
The actual moles of HNO3 used = 0.147 mol
The moles of HNO3 needed = 0.027375 mol
Since the moles of HNO3 needed (0.027375 mol) is smaller than the actual moles used (0.147 mol), HNO3 is the limiting reagent.
4. To find the excess reagent, subtract the moles of the limiting reagent from the total moles of the excess reagent.
Moles of excess HNO3 = Actual moles of HNO3 used - Moles of HNO3 needed
Moles of excess HNO3 = 0.147 mol - 0.027375 mol
= 0.119625 mol
5. Finally, calculate the mass of the excess HNO3 remaining using its molar mass:
Mass of HNO3 remaining = Moles of excess HNO3 * Molar mass of HNO3
Mass of HNO3 remaining = 0.119625 mol * 63.01 g/mol
= 7.532 g
Rounded to two decimal places, the mass of the excess reagent (HNO3) remaining after the complete reaction is 7.53 g.