Question

В 320 г 12%-го раствора бромида калия пропустили 56 мл газообразного хлора (н.у.). Сколько г нового простого вещества образовалось? Ответ дайте с точностью до десятых

Answer 1

M(KBr безв.)=170*0,12=20,4г
n(KBr безв.)=m/M=20,4/119=0,17моль
n(Cl2)=0,056/22,4=0,0025моль
m(Cl2)=n*M=0,0025*71=0,1175
n(KBr безв.)>n(Cl2)
изб. нед.
Считаем по хлору
0,1175г xг
2KBr+Cl2=2KCl+Br2
71 160г
х=0,1175*160/71=0,4 г
Масса брома 0,4 г

Answer 2

Calculation of the Amount of New Substance Formed

To calculate the amount of new substance formed in the reaction between a 12% solution of potassium bromide and 56 mL of chlorine gas, we need to use the concept of stoichiometry.

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. It allows us to determine the amount of one substance involved in a reaction based on the amount of another substance.

To solve this problem, we need to know the balanced chemical equation for the reaction between potassium bromide (KBr) and chlorine gas (Cl2). Unfortunately, the provided search results do not contain the necessary information to determine the balanced chemical equation for this specific reaction.

Without the balanced chemical equation, it is not possible to accurately calculate the amount of new substance formed. Therefore, I am unable to provide a precise answer to your question at this time.

If you have any other questions or need further assistance, feel free to ask!