If 2.00g of Zn is allowed to react with 1.75g of CuSO4, how many grams of Zn wil remain after the reaction is complete?
To find out how many grams of Zn will remain after the reaction, we first need to determine the limiting reactant.
Step 1: Calculate the molar mass of Zn and CuSO4.
- The molar mass of Zn is 65.38 g/mol.
- The molar mass of CuSO4 is 159.61 g/mol.
Step 2: Convert the given masses of Zn and CuSO4 to moles.
- Moles of Zn = mass / molar mass = 2.00g / 65.38 g/mol = 0.0305 mol
- Moles of CuSO4 = mass / molar mass = 1.75g / 159.61 g/mol = 0.0109 mol
Step 3: Determine the stoichiometric ratio between Zn and CuSO4 from the balanced equation.
- The equation for the reaction between Zn and CuSO4 is:
Zn + CuSO4 -> ZnSO4 + Cu
- From the balanced equation, we can see that 1 mole of Zn reacts with 1 mole of CuSO4.
Step 4: Identify the limiting reactant.
- The limiting reactant is the one that is completely consumed in the reaction. It can be determined by comparing the mole ratios of the reactants.
- The mole ratio of Zn to CuSO4 is 1:1.
- Since the moles of CuSO4 (0.0109 mol) are smaller than the moles of Zn (0.0305 mol), CuSO4 is the limiting reactant.
Step 5: Calculate the theoretical yield of ZnSO4 using the limiting reactant.
- The stoichiometric ratio of CuSO4 to ZnSO4 is 1:1.
- Therefore, the moles of ZnSO4 formed will also be 0.0109 mol.
Step 6: Calculate the mass of Zn that reacts.
- Moles of Zn used = moles of limiting reactant (CuSO4) x stoichiometric ratio (1 mol Zn / 1 mol CuSO4)
- Moles of Zn used = 0.0109 mol x 1 mol Zn / 1 mol CuSO4 = 0.0109 mol
- Mass of Zn used = moles of Zn used x molar mass of Zn = 0.0109 mol x 65.38 g/mol = 0.714 g (rounded to three decimal places)
Step 7: Calculate the remaining mass of Zn.
- Mass of Zn remaining = initial mass of Zn - mass of Zn used
- Mass of Zn remaining = 2.00 g - 0.714 g = 1.286 g (rounded to three decimal places)
Therefore, after the reaction is complete, 1.286 grams of Zn will remain.
To answer this question, we need to understand the reaction between zinc (Zn) and copper sulfate (CuSO4). The balanced chemical equation for this reaction is as follows:
Zn + CuSO4 → Cu + ZnSO4
From the balanced equation, we can determine that one mole of zinc (Zn) reacts with one mole of copper sulfate (CuSO4) to produce one mole of copper (Cu) and one mole of zinc sulfate (ZnSO4).
To calculate the amount of zinc remaining after the reaction, we need to follow these steps:
Step 1: Convert the given masses of Zn and CuSO4 into moles.
- The molar mass of Zn is 65.38 g/mol.
- The molar mass of CuSO4 is 159.61 g/mol.
Number of moles of Zn = 2.00 g / 65.38 g/mol
Number of moles of CuSO4 = 1.75 g / 159.61 g/mol
Step 2: Determine the limiting reactant.
- The limiting reactant is the one that is completely consumed in the reaction and determines the amount of product formed.
To find the limiting reactant, we need to compare the number of moles of Zn with the number of moles of CuSO4. The reactant with the smaller number of moles is the limiting reactant.
Step 3: Use stoichiometry to calculate the amount of zinc remaining.
- Since one mole of Zn reacts with one mole of CuSO4, the amount of zinc remaining will be equal to the initial amount of zinc minus the amount of zinc that reacted.
Amount of zinc remaining = Initial amount of Zn - Amount of Zn reacted
Step 4: Calculate the amount of zinc reacted.
- Since the balanced equation shows a 1:1 mole ratio between Zn and CuSO4, the moles of zinc reacted will be equal to the moles of copper sulfate.
Amount of Zn reacted = Number of moles of CuSO4
Finally, we can substitute the values and calculate the amount of zinc remaining.
Please provide the masses of Zn and CuSO4 to proceed with the calculations.
This is a limiting reagent problem. How do I know that? Because BOTH reactants are given. Basically, you work one of these problems by solving two simple stoichiometry problems in one; that is, you determine the product with one of the reactants (ignore the other one), then switch reactants and ignore the other one. The answers you get for the product are different and the smaller one is ALWAYS the correct answer. Here is how you do the simple stoichiometry problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html