How many milliliters of 3.00 M NaOH are required to react with 4.00 mL of 16.0 M HNO3?
How many milliliters of 3.00 M NaOH are required to react with 0.500 g of Cu+2 to form Cu(OH)2?
Based on your answers, question 2 and 3, if a student added 25mL of NaOH, in part 2 of the experiment, there will be enough NaOH to react with both the HNO3 and Cu+2.True False
A student decomposed 3.67g of copper (II) hydroxide into copper (II) oxide. How many mL of 3M H2SO4 is need to react with all the copper (II) oxide?
Using the information in question 5, how much copper (II) sulfate, in grams, are prodcued?
Using the amount of copper (II) sulfate produced, in question 6, how much copper (in grams) is produced if excess zinc is used?
In step 5 , of the experiment, 6M HCl is added to react with any unreactive Zn left in solution. If a student has 5.21g of copper (II) sulfate, how much zinc is needed to react with copper so that no HCl is needed in this step?
1. For the first question, we can use the formula: M1V1 = M2V2.
V2 = (M1V1) / M2
V2 = (3.00 mol/L * 4.00 mL) / 16.0 mol/L
V2 = 0.75 mL
2. For the second question, we first need to calculate the moles of Cu+2 using its molar mass.
Moles of Cu+2 = 0.500g / 63.55g/mol = 0.00787 mol
Now, we can use the formula: M1V1 = M2V2.
V2 = (M1V1) / M2
V2 = (3.00 mol/L * 0.00787 mol) / 1.00 mol/L
V2 = 0.0236 mL
3. True. Since the total volume required from both reactions is 0.75 mL + 0.0236 mL = 0.7736 mL, which is less than the 25 mL of NaOH added.
4. First, we need to calculate the moles of CuO using its molar mass.
Moles of CuO = 3.67g / 79.55g/mol = 0.0461 mol
Now we use the formula: M1V1 = M2V2.
V1 = (0.0461 mol * 1 L / 3M) * 1000 mL/L
V1 = 15.37 mL
5. Using stoichiometry, we can determine the moles of CuSO4 produced from the moles of CuO reacted with H2SO4.
From the balanced chemical equation: CuO + H2SO4 -> CuSO4 + H2O
1 mole of CuO produces 1 mole of CuSO4
Therefore, 0.0461 mol of CuO produces 0.0461 mol of CuSO4.
Mass of CuSO4 = 0.0461 mol * 159.61 g/mol = 7.36 g
6. Using stoichiometry again, we can determine the moles of Cu produced from the moles of CuSO4 reacted with excess zinc.
From the balanced chemical equation: CuSO4 + Zn -> ZnSO4 + Cu
1 mole of CuSO4 produces 1 mole of Cu
Therefore, 0.0461 mol of CuSO4 produces 0.0461 mol of Cu.
Mass of Cu = 0.0461 mol * 63.55 g/mol = 2.93 g
7. If 5.21g of CuSO4 is obtained and all of it comes from 0.0461 mol of CuO, then we need to calculate the moles of Cu needed to produce that amount of CuSO4.
Moles of Cu = 5.21g / 159.61g/mol = 0.0326 mol
From stoichiometry, 1 mole of Cu produces 1 mole of CuSO4.
Therefore, 0.0326 mol of Cu produces 0.0326 mol of CuSO4.
This means that 0.0326 mol of CuO is needed to produce 5.21g of CuSO4.
Now we can calculate the grams of CuO needed for the reaction.
Mass of CuO = 0.0326 mol * 79.55 g/mol = 2.59 g
Therefore, 2.59 grams of CuO is needed to produce 5.21g of CuSO4.