what mass of nitrogen(v)oxide would be obtained by heating 33.3g of lead trioxonitrate(v)

I need the final answer for this question

I need the final answer

You need to learn how to name compounds. I assume nitrogen oxide(v) means N2O5; however, when lead(II) nitrate is heated this is what is produced.

2Pb(NO3)2 ==> 2PbO + O2 + 4NO2
mols Pb(NO3)2 = grams/molar mass = estimated 0.1
For every 2 mols Pb(NO3)2 you get 4 mol NO2 so for estd 0.1 mol Pb(NO3)2 you get estd 0.2 mol NO2.
Then mass NO2 = mols NO2 x molar mass NO2.
Also, there are several names for Pb(NO3)2 but lead trioxonitrate(v) is not one of them.

Oh, heating things up, are we? Well, let me tell you, chemistry can be quite explosive! Now, to answer your question, when you heat 33.3g of lead trioxonitrate(V), you'll be left with a mass of nitrogen(V) oxide. Considering its chemical formula, I'll need to calculate the molar mass of lead trioxonitrate(V), which is approximately 331.2 g/mol. Now, I must warn you, the actual amount of nitrogen(V) oxide obtained may vary due to experimental conditions and ventilation. Don't want any dangerous fumes floating around!

To determine the mass of nitrogen(V) oxide produced by heating lead trioxonitrate(V), we need to balance the chemical equation.

The chemical formula for lead trioxonitrate(V) is Pb(NO3)2, and when heated, it decomposes into lead(II) oxide (PbO), nitrogen(IV) oxide (NO2), and oxygen gas (O2).

The balanced chemical equation for this reaction is:

2Pb(NO3)2 → 2PbO + 4NO2 + O2

From the balanced equation, we can see that for every 2 moles of lead trioxonitrate(V) decomposed, we obtain 4 moles of nitrogen(IV) oxide.

First, let's convert the given mass of lead trioxonitrate(V) (33.3g) into moles:

Using the molar mass of Pb(NO3)2, which is approximately 331.21 g/mol:
33.3g Pb(NO3)2 * (1 mol / 331.21 g) ≈ 0.1005 mol Pb(NO3)2

According to the balanced chemical equation, for every 2 moles of Pb(NO3)2, we obtain 4 moles of NO2.

So, we can calculate the moles of NO2 produced:
0.1005mol Pb(NO3)2 * (4 mol NO2 / 2 mol Pb(NO3)2) = 0.201 mol NO2

Finally, let's convert the moles of NO2 into grams, using the molar mass of NO2, which is approximately 46.01 g/mol:
0.201 mol NO2 * 46.01 g/mol ≈ 9.255 g NO2

Therefore, heating 33.3g of lead trioxonitrate(V) would produce approximately 9.255 grams of nitrogen(IV) oxide.

Yes