Balance your chemical reaction to ensure that the quantities on the left side equal those on the right side. For example, in the equation "Zn(NO3)2 --->ZnO + NO2 + O2," you have six oxygens on the left side, but only four on the right. To remedy this, you can add a "2" coefficient in front of "ZnO." Then, you have two zincs on the right side and only one on the left side. Add a "2" coefficient in front of "Zn(NO3)2." Now, you have four nitrogens on the left side and only one on the left. Add a "4" coefficient in front of "NO2." Now, your equation is balanced--you have two zincs, four nitrogens and twelve oxygens on either side.
In this instance, your balanced reaction will read: 2Zn(NO3)2 ---> 2ZnO + 4NO2 + O2. In other words, for every two moles of zinc nitrate you burn, you will yield two moles of zinc oxide, four moles of nitrogen dioxide and one mole of molecular oxygen.
Compute your reactants and products' molecular weights using your periodic table. For zinc nitrate, keep in mind that you need to add together the weights of one zinc atom, two nitrate atoms and six oxygen atoms to yield 189.36 atomic mass units (amu). You should get molecular weights of 81.38 amu, 46.00 amu and 32.00 amu for zinc oxide, nitrogen dioxide and molecular oxygen, respectively.
Determine how many moles of reactant you have, keeping in mind that one mole is equal to the same mass of substance (in grams) as the molecular weight in amu. For example, one mole of zinc nitrate is equal to 189.36 grams. Imagine that you have 300 grams of zinc nitrate. Complete the following conversion: 300 g Zn(NO2)3 x (1 mol Zn(NO2)3/189.36 g Zn(NO2)3) = 1.58 mol Zn(NO2)3.
Calculate how many moles of each product you can expect after your reaction, keeping in mind the ratios the balanced reaction "2Zn(NO3)2 --> 2ZnO + 4NO2 + O2" establishes. For example, because zinc oxide also has a "2" coefficient in front of it, you can expect exactly the same number of moles (1.58) of ZnO as you had of Zn(NO2)3. On the other hand, the reaction will yield two moles of nitrogen dioxide and one-half mole of molecular oxygen for each mole of zinc nitrate burned, or 3.16 mol NO2 and 0.79 mol O2, respectively.
Convert these molar amounts to masses using the molecular weights you determined. For example, you can convert 1.58 moles of zinc oxide using this conversion factor: 1.58 mol ZnO x (81.38 g ZnO/1 mol ZnO) = 128.58 g ZnO. Ultimately, you should determine that the thermal decomposition of 300 g Zn(NO3)2 will yield 145.36 g NO2 and 25.28 g of O2, in addition to 128.58 g ZnO.