Question

What is the mass of Carbon present in 0.5 mole of Pottasium Ferrocynide ( K4[Fe(CN)6] )

Answer 1

Total molar mass of K4[Fe(CN)6] = 368 g/mol
Mass of 1 mol of K4[Fe(CN)6] = 368 g/mol
So, Mass of 0.5 mol of K4[Fe(CN)6] = 368 x 0.5 g/mol = 184 g/mol
Mass of C in K4[Fe(CN)6]= 6x 12=72 g/mol
Let mass of Carbon in this compound is A
Then, A/184 = 72/368 = 36 g/mol
hence, mass of carbon in 0.5 mol is 36 g/mol

Answer 2

Total molar mass of K4[Fe(CN)6] = 368 g/mol

Mass of 1 mol of K4[Fe(CN)6] = 368 g/mol

So, Mass of 0.5 mol of K4[Fe(CN)6] = 368 x 0.5 g/mol = 184 g/mol

Mass of C in K4[Fe(CN)6]= 6x 12=72 g/mol

Let mass of Carbon in this compound is A

Then, A/184 = 72/368 = 36 g/mol

hence, mass of carbon in 0.5 mol is 36 g/mol

Read more on Brainly.in - https://brainly.in/question/333072#readmore