How many {\rm NO}_{2} molecules would form as products if the reaction had a percent yield of 75\%?
6 Mols of NO_2
This is due to there being 8 mols in the theoritical yeild and using the % yeild formula.
Sorry I'm not too sure how to explain how I did this.
To determine the number of NO2 molecules that would form as products with a percent yield of 75%, we need to know the amount of the reactant that is available. Without that information, we won't be able to calculate the number of molecules.
However, I can explain the general procedure you would follow to calculate the number of molecules based on the percent yield.
1. Start by determining the balanced chemical equation for the reaction. In this case, let's say the reaction is as follows:
2 NO + O2 -> 2 NO2
2. Identify the stoichiometry of the reactants and products. From the balanced equation, we can see that 2 moles of NO react with 1 mole of O2 to produce 2 moles of NO2.
3. Convert the given quantity of the reactant into moles. Suppose you have the number of moles of NO available.
4. Use the stoichiometry from the balanced equation to determine the number of moles of NO2 that should be formed. Multiply the number of moles of NO by the stoichiometric coefficient ratio. In this case, you would multiply the moles of NO by 2 to get moles of NO2.
5. Finally, apply the percent yield. Multiply the calculated moles of NO2 from step 4 by the percent yield (in decimal form) to get the actual moles of NO2 produced.
Please provide the amount of NO available or any other necessary information to proceed with the calculation.