Question

Answer 1

4K+O₂-to>2K₂O

0,1435---------0,07175

n _K=\(\dfrac{5,6}{39}\)=0,1435 mol

n _O2=\(\dfrac{11,2}{22,4}\)=0,5 mol

=>O2 dư

=>m _cr=0,07175.94+21,6=28,3445g

Answer 2

bài 3

2Cu+O₂-to>2CuO

0,2----------------0,2 mol

n _Cu=\(\dfrac{12,8}{64}\)=0,2 mol

=>dựa theo bài 2 =>Cu hết , oxi dư

=>m _cr=28,3445+0,2.80=44,3445g