Question

Report your answer to 3 significant figures in scientific notation (example: 1.236+6) unless stated otherwise. DO NOT include the units in your answer. 1. Consider a reaction of 744 mL of a 1.08x102 M Na3A9(503) solution completely reacts with 251 L 0 4.55*10-2 M Naer solution according to the following UNBALANCED chemical equation, NA (S203)2(aq) + NaBr(aq) -- Agr(s) + Nay5303() (a) Balance the chemical equation using the lowest integer coefficients. Be sure to include all coefficients including those of 1. NasAg(S0() NaBrag) - Ager() Na $20(4) Dutt Aww Tries 0/99 (b) Determine the mass (ing) of AgBr that is formed in the reaction Report your answer to 3 significant figures in standard notation (example: 1.23). 9 Tries 0/99 mit An ( 1.25 g of Agris produced, what is the percent yield of the reaction Report your answer to 3 significant figures in standard notation (example: 1.25). Tries 0/99 2. (6) How many moles of Catlons are present in 2.20x10? mL of a 2.26x102 M solution of Caciosa? 2. (a) How many moles of Conlons are present in 2,20x10mt of a 2.26x102M solution of Ca(C103)? mol Tries 0/99 Subint Anne (b) What volume of a 5.93x10-4 M solution of NICCIO4)2 (in L.) contains 3.61*10*1 moles of Cioton? L Submit Answer Tries 0/99 (s) Calculate the molarity (in mol/L) of Fions present in an aqueous solution prepared by adding 744 ml of 6.86x101 M aqueous Fef, to 251 ml of water. mol/L Tries 0/99

Answer 1

To balance the chemical equation, use stoichiometry to determine the mass of AgBr formed, and calculate the percent yield of the reaction.

Step-by-step explanation:

To balance the chemical equation Na2SO3 + NaBr -> NaBr + Na2SO4, the balanced equation is 2NaBr + SO2 + 2H2O -> 2NaHSO3. This means that for every 2 moles of NaBr and SO2, 2 moles of NaHSO3 are formed.

To determine the mass of AgBr formed in the reaction, we need to use stoichiometry. The balanced equation is 2AgNO3 + NaBr -> 2AgBr + NaNO3. From this equation, we can see that 2 moles of AgBr are formed for every 2 moles of AgNO3. We can use the molar mass of AgBr to convert moles to grams.

The percent yield of the reaction can be calculated by dividing the actual yield (given as 1.25 g) by the theoretical yield (calculated from stoichiometry) and multiplying by 100%.