I solve these limiting reagent problems the long way by solving two simpler stoichiometry problems. Here is a step by step procedure for solving the simple stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Solve the first time using moles NaI (moles = M x L) and solve for moles I2 produced. Solve the second time using moles Br2 for moles I2 formed. You have two answers and of course one of them must be wrong. In limiting reagent problems, the smaller value is ALWAYS the correct one to use.
The equation is
2NaI + Br2 ==> 2NaBr + I2.
After finding the theoretical yield from the calculations above, then
theoretical yield x 0.81 = actual yield.
Post your work if you get stuck.
Find the theoretical yield when 100 mL of 0.1 M NaI is reacted with 0.50 mL of Br2 (density: 3.10 g/mL) to produce NaBr and I2. What is the actual yield if the percent yield is 81%? (Atomic masses (g/mol): Na, 22.99; I, 126.9; Br, 79.90)
1 answer