a, Ta co pthh
SO3 + H2O \(\rightarrow\)H2SO4
Theo de bai ta co
nSO3=\(\dfrac{8}{80}=0,1mol\)
500ml=0,5 l
Theo pthh
nH2SO4=nSO3=0,1 mol
\(\Rightarrow\)Nong do mol cua dd la
CM= \(\dfrac{n}{v}=\)\(\dfrac{0,1}{0,5}=0,2M\)
b, CuO + H2SO4\(\rightarrow\)CuSO4 + H2O
Theo de bai ta co
nCuO=\(\dfrac{10}{80}=0,125mol\)
Nong do mol cua chat sau phan ung la
CM=\(\dfrac{n}{v}=\dfrac{0,125}{0,5}=0,25M\)
a/nSO3=8/79=0,1 (mol)
ADTC :CM=n/V
CM=0,1/0,5=0,2(M)
-nCuO=10/80=0,125( mol)
ADCT:CM=n/V
CMH2SO4=0,125/0,5=0,25(M)