quiz × bcswebdav/pid-4251246-dt-content-rid-28528217 Updates Not som updates aut
ID: 541487 • Letter: Q
Question
Explanation / Answer
The balanced equation
5NaClO3 + 3I2 + 3H2O ---------------> 5NaCl + 6HIO3
no of moles of NaClO3 = molarity * volume in L
= 0.0833*0.025 = 0.00208 moles
5 moles of NaClO3 react with 3 moles of I2
0.0021 moles of NaClO3 react with = 3*0.00208/5 = 0.00125moles of I2
mass of I2 = no of moles * gram molar mass
= 0.00125*253.8 =0.317g
moles NaClO3 = 25 x 0.0833 /1000 = 0.00208
The ratio between NaClO3 and I2 is 5 : 3
5 : 3 = 0.00208 : x
x = 0.00125 moles I2
Molecular weight of I2 = 253.8 g/mol
253.8 g/mol x 0.00125 mol = 0.317 g of I2
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