A7.0g sample of cu(no3)2•nh20 is heated, and 4.3g of anhydrous salt remains. what is the value of n

6.5  

Step-by-step explanation:

We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:                                        187.56     18.02

           Cu(NO₃)₂·nH₂O ⟶ Cu(NO₃)₂ + nH₂O

m/g:             7.0                        4.3

1. Moles of Cu(NO₃)₂

Moles of Cu(NO₃)₂ = 4.3 g × (1 mol/187.56 g)

Moles of Cu(NO₃)₂ = 0.0229 mol

2. Mass of H₂O

Mass of Cu(NO₃)₂·nH₂O = mass of Cu(NO₃)₂ + mass of H₂O

                 7.0                 =               4.3           +           x

                 7.0 -  4.3        =                                            x

                 2.7                 =                                             x

3. Moles of H₂O

Moles of H₂O = 2.7 g × (1 mol/18.02 g)

Moles of H₂O = 0.150 mol

4. Value of n

The molar ratio is 1 mol (NO₃)₂ = n mol H₂O

n = moles H₂O/moles Cu(NO₃)2

n = 0.150/0.0229

n = 6.5

This answer does not make sense, because the maximum value of n in hydrated copper(II) nitrate is 6.

- lead + phosphoric acid = hydrogen + lead(ii) phosphate