if 2.00g of Zn is allowed to react with 2.00g of CuSO4 according to the equation CuSO4(aq) + Zn(s)-----> ZnSO4 (aq)+ Cu(s)

how many grams of Zn will remain after the reaction is complete ?

Well, let's see. First, we need to calculate the number of moles of Zn and CuSO4.

For Zn:
Molar mass of Zn = 65.38 g/mol
Number of moles of Zn = 2.00g / 65.38 g/mol = 0.0306 mol

For CuSO4:
Molar mass of CuSO4 = 159.61 g/mol
Number of moles of CuSO4 = 2.00g / 159.61 g/mol = 0.0125 mol

According to the balanced equation, the ratio of Zn to CuSO4 is 1:1. So, the limiting reactant is CuSO4. This means all the CuSO4 will be consumed in the reaction, and Zn would be left over.

So, to determine the grams of Zn remaining, we need to calculate the grams of Zn reacted with CuSO4.

Molar mass of Zn = 65.38 g/mol
Number of moles of Zn reacted = 0.0125 mol

Grams of Zn reacted = 65.38 g/mol × 0.0125 mol = 0.817 g

To find the grams of Zn remaining, we subtract the grams of Zn reacted from the initial grams of Zn.

Grams of Zn remaining = 2.00 g - 0.817 g = 1.183 g

So, 1.183 grams of Zn will remain after the reaction is complete.

To find out how many grams of Zn will remain after the reaction is complete, we need to determine the limiting reactant first.

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 into moles.
- Moles of Zn = Mass of Zn / Molar mass of Zn
= 2.00 g / 65.38 g/mol
≈ 0.03055 mol

- Moles of CuSO4 = Mass of CuSO4 / Molar mass of CuSO4
= 2.00 g / 159.61 g/mol
≈ 0.01253 mol

Step 3: Use the stoichiometric coefficients in the balanced equation to determine the limiting reactant.

From the equation CuSO4(aq) + Zn(s) → ZnSO4(aq) + Cu(s), we can see that the stoichiometric ratio between CuSO4 and Zn is 1:1. Therefore, the limiting reactant is the one with fewer moles.

In this case, CuSO4 has fewer moles (0.01253 mol) compared to Zn (0.03055 mol). So, CuSO4 is the limiting reactant.

Step 4: Calculate the moles of Zn that reacted with CuSO4.
- Based on the stoichiometry, 1 mole of CuSO4 reacts with 1 mole of Zn.
- Therefore, the moles of Zn that reacted = 0.01253 mol

Step 5: Calculate the remaining moles of Zn.
- Remaining moles of Zn = Total moles of Zn - Moles of Zn reacted
= 0.03055 mol - 0.01253 mol
≈ 0.01802 mol

Step 6: Convert the remaining moles of Zn into grams.
- Remaining mass of Zn = Remaining moles of Zn × Molar mass of Zn
≈ 0.01802 mol × 65.38 g/mol
≈ 1.18 g

Therefore, after the reaction is complete, approximately 1.18 grams of Zn will remain.

To determine how many grams of Zn will remain after the reaction is complete, we need to calculate the amount of Zn that reacts with CuSO4 and then subtract it from the initial amount.

1. Determine the molar masses:
- Molar mass of Zn (Zinc) = 65.38 g/mol
- Molar mass of CuSO4 (Copper(II) sulfate) = 159.61 g/mol

2. Convert the given masses to moles using the molar masses:
- Moles of Zn = 2.00 g / 65.38 g/mol = 0.0306 mol
- Moles of CuSO4 = 2.00 g / 159.61 g/mol = 0.0125 mol

3. Determine the stoichiometric ratio between Zn and CuSO4 from the balanced equation:
- From the equation CuSO4 + Zn -> ZnSO4 + Cu, we can see that the ratio between Zn and CuSO4 is 1:1.

4. Since the stoichiometric ratio is 1:1, the moles of Zn that react with CuSO4 will be equal to the moles of CuSO4:
- Moles of Zn reacted = Moles of CuSO4 = 0.0125 mol

5. Subtract the moles of Zn reacted from the initial moles of Zn to find the remaining moles:
- Moles of Zn remaining = Moles of Zn - Moles of Zn reacted = 0.0306 mol - 0.0125 mol = 0.0181 mol

6. Finally, convert the moles of Zn remaining to grams using the molar mass:
- Grams of Zn remaining = Moles of Zn remaining * Molar mass of Zn = 0.0181 mol * 65.38 g/mol = 1.18 g

Therefore, after the reaction is complete, approximately 1.18 grams of Zn will remain.

explain the orgin of X-rays

This is a limiting reagent problem. I know that because amounts are given for BOTH reactants. However since the problem asks for the amount of Zn remaining, one can assume that CuSO4 is the limiting reagent. I will show how we can confirm that.

CuSO4(aq) + Zn ==> ZnSO4(aq) + Cu(s)
mols CuSO4 = g/molar mass = 2.00/about 159.6 = 0.125

mols Zn = 2.00/65.4 = 0.0306

Convert mols of each to mols Cu produced.
0.0125 mols CuSO4 x (1 mol Cu/1 mol CuSO4) = 0.0125 mols Cu
0.0306 mols Zn x (1 mol Cu/1 mol Zn) = 0.0306 mols Cu
Of course both answers can't be correct; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that number is the limiting reagent. Therefore, 0.0125 mols Cu will be formed and CuSO4 is the limiting reagent. All of this just confirms what we knew at the beginning.

How much Zn remains unreacted? How many mols Zn were used? That will be
0.0125 mols Cu x (1 mol Zn/1 mol Cu) = 0.0125 mols Zn used.
mols Zn remaining = mols Zn initially-mols Zn used = 0.0306-0.0125 = ?
You can convert to grams by g Zn = mols Zn x atomic mass Zn.
mols Zn x (1 mul