Assuming complete reaction, what volume of 0.200 mol dm–3 potassium hydroxide solution (koh( is required to neutralize 25.0 cm3 of 0.200 mol dm–3 aqueous sulfuric acid, (h2so4(aq))

50cm^3

Explanation:

Step 1:

We'll begin by writing a balanced equation for the reaction. This is illustrated below:

2KOH + H2SO4 —> K2SO4 + 2H2O

From the balanced equation above, we obtained:

Mole of the acid (nA) = 1

Mole of the base (nB) = 2

Step 2:

Data obtained from the question include:

Concentration of the base, KOH (Cb) = 0.2moldm^-3

Volume of the base, KOH (Vb) =?

Volume of the acid, H2SO4 (Va) = 25cm^3

Concentration of the acid, H2SO4 (Ca) = 0.2moldm^-3

Step 3:

Determination of the volume of potassium hydroxide (KOH) needed for the complete reaction. This is illustrated below:

Applying the formula CaVa/CbVb = nA/nB, the volume of the base can be obtained as follow:

CaVa/CbVb = nA/nB

0.2 x 25/ 0.2 x Vb = 1/2

Cross multiply to express in linear form as shown below:

0.2 x Vb = 0.2 x 25 x 2

Divide both side by 0.2

Vb = (0.2 x 25 x 2)/0.2

Vb = 50cm^3

Therefore, the volume of the base, KOH required for the complete reaction is 50cm^3

not clear

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%