Answer:
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
