1. a pyrotechnician mixes 280 g of black powder for a firework. find the amount of oxygen, carbon dioxide, and sulfur dioxide generated when this firework bursts, if the oxidizing agent in the powder is saltpeter. 2. which are the limiting reactants for each reaction? based on this, calculate the maximum yield of oxygen, carbon dioxide, and sulfur dioxide when 280 g of black powder is burned. 3. when the pyrotechnician tested the firework, he found the actual yields of oxygen, carbon dioxide, and sulfur dioxide to be 81.57 g, 82.75 g, and 46.5 g respectively. 4. what is the percent yield of each gas? why should pyrotechnicians worry about the yield of the reactants in a firework? how does the yield affect the firework and the environment?

There are several issues with this question, as it does not provide the ratios of starting materials that make up the black powder, and do not provide proper products.  the actual reaction is as follows: 10kno₃ + 8c + 3s  → 2k₂co₃ + 3k₂so₄ + 6co₂ + 5n₂as we can see, oxygen and sulfur dioxide gases are not products of this reaction. also, the make up of black powder is roughly 75% kno₃, 15% c, 10% s by mass. we can use this information to answer these questions.1.) we can only find the amount of carbon dioxide formed since neither oxygen or sulfur dioxide are products. therefore, 0g of each are formed.280 g black powder x 0.75 = 210 g kno₃ / 101.1 g/mol = 2.08 moles kno₃2.08 moles kno₃ x 6 moles co₂/10 moles kno₃ = 1.25 moles co₂1.25 moles co₂ x 44 g/mol = 55 g co₂therefore, only 55 g of co₂ is formed.2.) we already determined that using the kno₃ we get 1.25 moles of co₂. we can do this calculation twice more using c and s as the starting reactants.280 g x 0.15 = 42 g c/ 12 g/mol = 3.5 moles c x 6 moles co₂/8 moles c = 2.63 moles co₂280 g x 0.10 = 28 g / 32 g/mol = 0.875 moles s x 6 moles co₂/3 moles s = 1.75 moles co₂these are the maximum amounts of co₂ that could be formed from each reactant, however, kno₃ provides the least amount of co₂ and is therefore the limiting reagent.3.) it says he found 81.57 g o₂, 82.75 g co₂ and 46.5 g so₂. however, we already established that no oxygen or sulfur dioxide are formed. and we already calculated that only 55 g of co₂ can be produced by this reaction, so it is impossible to get 82.75 g of co₂. the only value that could apply were if we actually formed 46.5 g of co₂. if this were true we could calculate a percent yield as follows: (46.5 g co₂ / 55 g co₂) x 100% = 84.5 % yield co₂4.) pyrotechnicians need to be concerned with reactants in a firework because these reactants could accidentally combust if not handled carefully. the yield of the reactants affect the environment and the firework because the way the mixture is packaged can affect the behaviour. the more tightly packed the mixture is, the greater the rate of the combustion process since the confinements are more concentrated.

Most likely a is the answer - because  you generate   38 atp's but also consume an atp.   each atp gives approx. less than 4 kcal so probably a is the closest answer.