Question

При взаимодействии бромида калия с хлором получили 16г брома.Вычислите массу взятого реакции бромида калия.Помогите решить плис!

Answer 1

Составим уравнение реакции:

2KBr+CL2=Br2+2KCl

n(Br2)=m/M=16г/160=0,1 моль

Переносим кол-во вещества, согласно поставленным коэффицентам.

т.е n(KBr)=0.2 моль

m(KBr)=n*M=0,2 моль* 120 г/моль=24г

 

Ответ: 24г.

Answer 2

Calculation of the Mass of Potassium Bromide

To calculate the mass of potassium bromide (KBr) used in the reaction, we need to use the given information that 16 grams of bromine (Br) were obtained.

The balanced chemical equation for the reaction between potassium bromide and chlorine is:

2 KBr + Cl2 → 2 KCl + Br2

From the equation, we can see that 2 moles of potassium bromide react to produce 1 mole of bromine. We can use this information to calculate the molar mass of potassium bromide and then determine the mass of potassium bromide used.

The molar mass of potassium bromide (KBr) can be calculated as follows:

Molar mass of KBr = Molar mass of K + Molar mass of Br = (39.10 g/mol) + (79.90 g/mol) = 119.00 g/mol

Now, we can calculate the number of moles of bromine (Br2) produced using the given mass of 16 grams:

Number of moles of Br2 = Mass of Br2 / Molar mass of Br2 = 16 g / 159.80 g/mol ≈ 0.100 mol

Since 1 mole of bromine is produced from 2 moles of potassium bromide, we can calculate the number of moles of potassium bromide used:

Number of moles of KBr = (0.100 mol Br2) / (1 mol Br2 / 2 mol KBr) = 0.050 mol KBr

Finally, we can calculate the mass of potassium bromide used:

Mass of KBr = Number of moles of KBr × Molar mass of KBr = 0.050 mol × 119.00 g/mol = 5.95 g

Therefore, the mass of potassium bromide used in the reaction is approximately 5.95 grams.

Please note that the calculations provided are based on the information given and the balanced chemical equation.