A student determines the zinc content of a solution by first precipitating it as zinc hydroxide, and then decomposing the hydroxide to zinc oxide by heating. How many grams of zinc oxide should the student obtain if her solution contains 42.0 mL of 0.566 M zinc nitrate?

moles zinc nitrate=.042*.466 Zn(NO2)2

moles zinc= same number

Moles ZnO=same number

grams Zn oxiide= moles ZnO*formula mass.

To calculate the number of grams of zinc oxide obtained, we need to use stoichiometry. Let's break down the steps:

Step 1: Precipitation of zinc hydroxide
The balanced chemical equation for the precipitation reaction is:
Zn(NO3)2 + 2NaOH → Zn(OH)2 + 2NaNO3

From the balanced equation, we can see that for every 1 mole of Zn(NO3)2, we obtain 1 mole of Zn(OH)2. Therefore, 1 mole of Zn(OH)2 is equal to the molar mass of Zn(OH)2.

Step 2: Decomposition of zinc hydroxide to zinc oxide by heating
The balanced chemical equation for the decomposition reaction is:
Zn(OH)2 → ZnO + H2O

From the balanced equation, we can see that for every 1 mole of Zn(OH)2, we obtain 1 mole of ZnO.

Now let's calculate the number of moles of zinc nitrate (Zn(NO3)2) in the solution:
moles of Zn(NO3)2 = concentration (M) × volume (L)
moles of Zn(NO3)2 = 0.566 M × 0.0420 L

Next, using the stoichiometry from the balanced equation, we can determine the number of moles of Zn(OH)2:
moles of Zn(OH)2 = moles of Zn(NO3)2

Finally, we can calculate the number of grams of ZnO using the molar mass of ZnO:
grams of ZnO = moles of Zn(OH)2 × molar mass of ZnO

Let's plug in the values to calculate the grams of ZnO obtained.

To determine the amount of zinc oxide obtained, we need to follow the stoichiometry of the reaction and use the given information.

First, let's write the balanced chemical equation for the reaction. The precipitate zinc hydroxide reacts with heat to form zinc oxide, as follows:

Zn(OH)2(s) → ZnO(s) + H2O(g)

From the equation, we see that 1 mole of zinc hydroxide will produce 1 mole of zinc oxide.

Next, we need to find the number of moles of zinc nitrate in the given solution. We can use the formula:

moles = concentration (M) × volume (L)

Given:
- Volume of the zinc nitrate solution = 42.0 mL = 0.0420 L
- Concentration of the zinc nitrate solution = 0.566 M

Using the formula, we can calculate the number of moles of zinc nitrate as follows:

moles of zinc nitrate = 0.566 M × 0.0420 L = 0.0238 moles

Since zinc hydroxide has a 1:1 stoichiometric relationship with zinc oxide, the number of moles of zinc oxide formed will be the same as the number of moles of zinc nitrate.

Therefore, the student should obtain 0.0238 moles of zinc oxide.

Finally, we can convert the moles of zinc oxide to grams using its molar mass. The molar mass of zinc oxide (ZnO) is 81.38 g/mol.

grams of zinc oxide = moles of zinc oxide × molar mass of zinc oxide

grams of zinc oxide = 0.0238 mol × 81.38 g/mol ≈ 1.94 grams

So, the student should obtain approximately 1.94 grams of zinc oxide.

mols Zn(NO3)2 = M x L = ?

You can get 1 mol Zn(OH)2 from 1 mol Zn(NO3)2; therefore mol Zn(OH)2 = mols Zn(NO3)2.
The same reasoning tells you that you can obtain the same number of mols ZnO.
g ZnO = mols x molar mass.