An excess of sodium hydroxide solution was added to 240 g of a solution g by weight of iron (II) nitrate

An excess of sodium hydroxide solution was added to 240 g of a solution g by weight of iron (II) nitrate 2%. Determine the mass of the precipitated sediment.

Given:

m (Fe (NO3) 3) = 240 g

w% (Fe (NO3) 3) = 2%

To find:

m (draft) -?

Decision:

Fe (NO3) 3 + 3NaOH = Fe (OH) 3 + 3NaNO3, – we solve the problem, relying on the composed reaction equation:

1) Find the mass of iron nitrate in solution:

m (Fe (NO3) 3) = 240 g * 0.02 = 4.8 g

2) Find the amount of iron nitrate:

n (Fe (NO3) 3) = m: M = 4.8 g: 242 g / mol = 0.02 mol

3) We compose a logical expression:

if 1 mol of Fe (NO3) 3 gives 1 mol of Fe (OH) 3,

then 0.02 mol Fe (NO3) 3 will give x mol Fe (OH) 3,

then x = 0.02 mol.

4) Find the mass of precipitated iron hydroxide:

m (Fe (OH) 3) = n * M = 0.02 mol * 107 g / mol = 2.14 g.

Answer: m (Fe (OH) 3) = 2.14 g.