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

answer: -

reaction:

na + cl ⇌ nacl

balanced chemical equation:

na + 2cl ⇌ nacl2

sodium + clorine goes to form sodium chloride (table salt).

the product is nacl2.

and the reactants are na and cl.

(note: if you don’t know what ⇌ means, read below: -

⇌(double head arrow) means reversible reaction).

c

explanation: