Question

cho Al tac dung voi 150 ml dd h2so4 2M . tinh mAl , tinh H2 va Tinh Cm cua dd sau pu biet Vdd thay doi ko dang ke

Answer 1

buithianhthoDuong LeQuang Nhân

Answer 2

2Al +3H2SO4---.Al2(SO4)3 +3H2

a)Ta có

n\(_{H2SO4}=0,15.2=0,3\left(mol\right)\)

Theo pthh

n\(_{Al}=\frac{2}{3}n_{H2SO4}=0,2\left(mol\right)\)

m\(_{Al}=0,2.27=5,4\left(g\right)\)

Theo pthh

n\(_{H2}=n_{H2SO4}=0,3\left(mol\right)\)

V\(_{H2}=0,3.22,4=6,72\left(l\right)\)

Theo pthh

n\(_{Al2\left(SO4\right)3}=\frac{1}{3}n_{H2SO4}=0,1\left(mol\right)\)

C\(_{\left(Al2\left(SO4\right)3\right)}=\frac{0,1}{0,15}=0,67M\)

Chúc bạn học tốt

Answer 3

https://i.imgur.com/fzDOnvy.jpg

Answer 4

PTHH:\(\text{2Al +3H2SO4--->Al2(SO4)3 +3H2}\)

a)Ta có:

nH2SO4=0,15.2=0,3(mol)

Theo pthh

\(\text{nAl=nH2SO4=0,2(mol)}\)

\(\text{m_{Al}=0,2.27=5,4(g)}\)

Theo pthh

nH2=nH2SO4=0,3(mol)

VH2=0,3.22,4=6,72(l)

Theo pthh

nAl2(SO4)3=13nH2SO4=0,1(mol)

C_M(Al2(SO4)3)=0,1/0,15=0,67M

#9A3