Answer:
13.9g of CuSO4•5H2O
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
CuSO4•5H2O → CuSO4 + 5H2O
Next, we shall determine the mass of CuSO4•5H2O heated and the mass of H2O produced from the balanced equation.
This is illustrated below
Molar mass of CuSO4• 5H2O = 63.5 + 32 + (16x4) + 5(2x1 + 16)
= 63.5 + 32 + 64 + 5(18) = 249.5g/mol
Mass of CuSO4•5H2O from the balanced equation = 1 x 249.5 = 249.5g
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H2O from the balanced equation = 5 x 18 = 90g
From the balanced equation above,
249.5g of CuSO4•5H2O produced 90g of H2O.
Now, we can determine the mass of CuSO4•5H2O needed to produce 5g of H2O. This can be achieved as shown below:
From the balanced equation above,
249.5g of CuSO4•5H2O produced 90g of H2O.
Therefore, Xg of CuSO4•5H2O will produce 5g of H2O i.e
Xg of CuSO4•5H2O = (249.5 x 5)/90
Xg of CuSO4•5H2O = 13.9g
Therefore, 13.9g of CuSO4•5H2O is needed to produce 5g of H2O.
![pH=-log[H^{+}]=-log[H_{3}O]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%7B%2B%7D%5D%3D-log%5BH_%7B3%7DO%5D%20%20)
. Using the value in formula, we get:![pH=-log[1.8*10^{-4}]=3.745](https://tex.z-dn.net/?f=pH%3D-log%5B1.8%2A10%5E%7B-4%7D%5D%3D3.745%20)