Question

Answer 1

1

nCuO=\(\dfrac{8}{80}=0,1mol\)

n Fe2(SO4)3=\(\dfrac{300}{400}=0,75mol\)

2. 

VH2=2.22,4=44,8l

n O2=\(\dfrac{16}{32}=0,5mol\)

=>VO2=0,5.22,4=11,2l

Answer 2

\(n_{CuO} = \dfrac{8}{80} = 0,1(mol)\\ n_{Fe_2(SO_4)_3} = \dfrac{300}{400} = 0,75(mol)\\ V_{H_2} = 2.22,4 = 44,8(lít)\\ n_{O_2} = \dfrac{16}{32} = 0,5(mol)\\ V_{O_2} = 0,5.22,4 = 11,2(lít)\)