Chemistry, asked by utkarshagarwal6258, 1 year ago

How many grams of potassium are present in 28.8 g of k2cr2o7?

Answers

Answered by priyabachala
19
molecular weight of k2cr2o7 is 294g
in 1mole of k2cr2o7 78grams of potassium is present
so in 28.8g of k2cr2o7 let x grams of potassium present
so
294g ------- 78g of potassium
28.8g -------- x g of potassium
by cross multiplication
294*x = 78*28.8
x = 7.64
so 7.64 grams of potassium is present

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