Determine the mass of 6.33 mol of iron(II) nitrate.

1822.72 g

Explanation:

Applying,

n = R.M/M.M Equation 1

Where n = number of moles of iron(II) nitrate, R.M = Reacting mass of iron(II) nitrate, M.M = molar mass of Iron(II) nitrate.

Make R.M the subject of the equation

R.M = n×M.M Equation 2

From the question,

Given: n = 6.33 mol

Constant: M.M of iron(II) nitrate = 287.95 g/mol

Substitute these values into equation 2

R.M = 6.33(287.95)

R.M = 1822.72 g

Hence the mass of iron(II) nitrate is 1822.72 g