The author posted a question in Homework
chemistry problem one more? and got a better answer
Response from Дима[+++]
1.58g KMnO4 - is 0.01mol 2.24l hydrogen sulfide - is 0.1mol By reaction to 1 mole of hydrogen sulfide we need 0.4 moles KMnO4, consequently permanganate in deficiency and calculation we conduct by it From 2 moles KMnO4 we get 5 moles of sulfur, and from 0.01mole - respectively 0.025 moles S proportion It will make 32*0.025=0.8g?
Response from 0[+++++]
1.58g KMnO4 is 0.01mol 2.24l hydrogen sulphide is 0.1mol According to reaction on 1 mol hydrogen sulphide it is necessary 0 4 mol KMnO4 consequently permanganate in lack and calculation we conduct on it From 2 mol KMnO4 is received 5 mol sulfur and from 0 01mol - respectively 0 025 mol S on a proportion it will make 32*0 025=0 8g.
1.58g KMnO4 is 0.01mol 2.24l hydrogen sulphide is 0.1mol According to reaction on 1 mol hydrogen sulphide it is necessary 0 4 mol KMnO4 consequently permanganate in lack and calculation we conduct on it From 2 mol KMnO4 is received 5 mol sulfur and from 0 01mol - respectively 0 025 mol S on a proportion it will make 32*0 025=0 8g.
Response from 0[++]
Well that's easy-
Well that's easy-