How to calculate this ?
Do you use the molar mass and multiply it by the mass given ?

Option C. 8.91g

Explanation:

The balanced equation for the reaction is given below:

2Pb(NO3)2 —> 2PbO + 4NO2 + O2

Next, we shall determine the mass of Pb(NO3)2 that decomposed and the mass PbO produced from the balanced equation.

This is illustrated below below:

Molar mass of Pb(NO3)2 = 207 + 2[14 + (16x3)]

= 207 + 2[14 + 48]

= 207 + 2[62] = 207 + 124

= 331g/mol

Mass of Pb(NO3)2 from the balanced equation = 2 x 331 = 662g

Molar mass of PbO = 207 + 16 = 223g/mol

Mass of PbO from the balanced equation = 2 x 223 = 446g

Summary:

From the balanced equation above,

662g of Pb(NO3)2 decomposed to produce 446g of PbO.

Finally, we can obtain the maximum mass of lead(II) oxide, PbO produced from 13.23g of lead (II) nitrate, Pb(NO3)2 as follow:

From the balanced equation above,

662g of Pb(NO3)2 decomposed to produce 446g of PbO.

Therefore, 13.23g of Pb(NO3)2 will decompose to produce = (13.23 x 446)/662 = 8.91g of PbO.

Therefore, 8.91g of PbO were obtained from the decomposition of 13.23g of Pb(NO3)2

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)