Answer to Question #81857 in General Chemistry for Megan

Question

How many grams of sodium iodide, NaI, must be used to produce 45.5 g of iodine, I2?

Answer 1

2NaI + 2KMnO4 + 2KOH → I2 + 2K2MnO4 + 2NaOH

M(I2) =254 g/mol

M(NaI) = 150 g/mol

m(I2) = 45,5 g

m(NaI) - ?

1. n(I2) = m(I2) / M(I2)

n(I2) = 45.5 g / 254 g/mol = 0.18 mol

2. n(NaI) =2 n(I2)

n(NaI) = 2*0.18 mol =0.36 mol

3. m(NaI) = n(NaI)/M(NaI)

m(NaI) = 0.36 mol*150 g/mol = 54 g