Question

Пожалуйста помогите!! очень срочно!! сколько граммов образуется при взаимодействии 160 г 20%-ного раствора cuso4 с необходимым количеством КОН

Answer 1

Объяснение:

m(CuSO4)=mp-pa•w(CuSO4)=160•0.2=32г

M (CuSO4) = 160 г/моль;

M Cu(OH)2 = 98 г/моль.

CuSO4 + 2KOH = K2SO4 + Cu(OH)2↓

n(CuSO4)=m/M=32/160=0.2моль

n(CuSO4)=n(Cu(ОН)2)=0.2моль

n(Cu(ОН)2)=n•M=0.2×98=19.6г

Answer 2

Calculation of Grams of CuSO4 Formed

To calculate the grams of CuSO4 formed when 160g of a 20% solution of CuSO4 reacts with the necessary amount of KOH, we can use the concept of stoichiometry and the given concentration of the CuSO4 solution.

First, we need to determine the amount of CuSO4 present in the 20% solution. The concentration of a solution is typically given in terms of grams of solute per 100 grams of solution. Therefore, a 20% solution of CuSO4 means there are 20g of CuSO4 in 100g of solution.

Given that the initial amount of the 20% CuSO4 solution is 160g, we can calculate the amount of CuSO4 present in the solution: Amount of CuSO4 in 160g of 20% solution = (20g/100g) * 160g

Next, we need to determine the stoichiometry of the reaction between CuSO4 and KOH to find the amount of CuSO4 that reacts with the necessary amount of KOH.

Let's calculate the amount of CuSO4 in the 160g of 20% solution and proceed with the stoichiometry calculation.

Now, let's calculate the amount of CuSO4 in the 160g of 20% solution.