The amount of CuSO₄.5H₂O require to prepare 6 g of copper ammine complex is 5.99 g.
Relation between mass and moles will be represented as:
n = W/M , where
W = given mass
M = molar mass
According to the question, formed product is copper ammine product i.e. [Cu(NH₃)]SO₄.H₂O from CuSO₄.5H₂O, i.e.
CuSO₄.5H₂O → [Cu(NH₃)₄]SO₄.H₂O
From the stoichiometry it is clear that 1 mole of given reactant is involved for the formation of 1 mole of product.
Now, moles of 6g of [Cu(NH₃)₄]SO₄.H₂O will be calculated as:
n = 6g / 245.79g/mol = 0.024 moles
So, 0.024 moles of CuSO₄.5H₂O is involved in this reaction and its weight will be calculated as:
W = (0.024mol)(249.69g/mol) = 5.99 g
Hence, required mass of the reactant is 5.99 g.
To know more about moles, visit the below link:
https://brainly.com/question/15374113
