Đáp án A.

đáp án a bạn nhé

vì nhôm pư với clo thep pthh:

2Al+3Cl₂\(\rightarrow\)2AlCl₃(1)

theo pthh(1) và đề bài ta có:nAlCl₃=26,7:133,5=0,2(mol)

nCl₂ =0,2:2\(\times\)3=0,3(mol)

mCl₂=71\(\times\)0,3=21,3(g)

vậ số gam clo cần dùng là 21,3(g)

Câu 1:

\(Mg+Br_2\rightarrow MgBr_2\\ n_{Br_2}=\dfrac{11,2}{160}=0,07\left(mol\right)=n_{Mg}=n_{MgBr_2}\\ a=m_{Mg}=0,07.24=1,68\left(g\right)\\ m_{MgBr_2}=184.0,07=12,88\left(g\right)\)

Mg+Br2->MgBr2

0,07--0,07----0,07

n Br2=\(\dfrac{11,2}{160}\)=0,07 mol

=>m Mg=0,07.24=1,68g

=>m MgBr2=0,07.184=12,88g

Câu 2:

\(n_{Al}=\dfrac{14,58}{27}=0,54\left(mol\right)\\ 2Al+3Cl_2\rightarrow\left(t^o\right)2AlCl_3\\ n_{AlCl_3}=n_{Al}=0,54\left(mol\right)\\ n_{Cl_2}=\dfrac{3}{2}.0,54=0,81\left(mol\right)\\ V_{Cl_2\left(đktc\right)}=0,81.22,4=18,144\left(l\right)\\ m_{AlCl_3}=0,54.133,5=72,09\left(g\right)\)

\(a,PTHH:Zn+Cl_2\rightarrow ZnCl_2\)

\(b,n_{Zn}=\dfrac{m}{M}=\dfrac{13}{65}=0,2\left(mol\right)\\ Theo.PTHH:n_{Cl_2}=n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\ a=m_{Cl_2}=n.M=0,4.35,5=14,2\left(g\right)\)

\(b=m_{ZnCl_2}=n.M=0,2.136=27,2\left(g\right)\)

\(c,PTHH:2Al+3Cl_2\rightarrow2AlCl_3\\ Theo.PTHH:n_{Al}=\dfrac{2}{3}.n_{Cl_2}=\dfrac{2}{3}.0,2=\dfrac{2}{15}\left(mol\right)\\ m_{Al}=n.M=\dfrac{2}{15}.27=3,6\left(g\right)\)

vì nhôm pư với clo thep pthh:

2Al+3Cl2$\to$2AlCl3(1)

theo pthh(1) và đề bài ta có:nAlCl3=26,7:133,5=0,2(mol)

nCl2 =0,2:2$\times$3=0,3(mol)

mCl2=71$\times$0,3=21,3(g)

vậy số gam clo cần dùng là 21,3(g)

PT: \(2Al+3Cl_2\underrightarrow{t^o}2AlCl_3\)

Ta có: \(n_{AlCl_3}=\dfrac{26,7}{133,5}=0,2\left(mol\right)\)

Theo PT: \(n_{Cl_2}=\dfrac{3}{2}n_{AlCl_3}=0,3\left(mol\right)\)

\(\Rightarrow m_{Cl_2}=0,3.71=21,3\left(g\right)\)

Bạn tham khảo nhé!

\(2Al + 3Cl_2 \xrightarrow{t^o} 2AlCl_3\\ n_{Cl_2} = \dfrac{3}{2}n_{AlCl_3} = \dfrac{3}{2}.\dfrac{26,7}{133,5} = 0,3(mol)\\ \Rightarrow m_{Cl_2} = 0,3.71 = 21,3(gam)\)

Đáp án A

Theo bài ra ta có:

56n_Fe + 64n_Cu =15,2

(Khối lượng ban đầu)

Lại có 3n_Fe + 2n_Cu = 0,3.2

(Bảo toàn e)

n_Fe = 0,1; n_Cu = 0,15

 %m_Fe = 36,84%; %m_Cu = 63,16%.

a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

b, \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)

\(n_{H_2}=\dfrac{3}{2}n_{Al}=0,45\left(mol\right)\Rightarrow V_{H_2}=0,45.22,4=10,08\left(l\right)\)

c, \(n_{AlCl_3}=n_{Al}=0,3\left(mol\right)\Rightarrow m_{AlCl_3}=0,3.133,5=40,05\left(g\right)\)

d, \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)

PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

Có: \(\dfrac{0,2}{1}>\dfrac{0,45}{3}\) → Fe₂O₃ dư.

\(n_{Fe}=\dfrac{2}{3}n_{H_2}=0,3\left(mol\right)\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(n_{Al}=\dfrac{3,375}{27}=0,125\)

\(\Rightarrow n_{H_2}=\dfrac{0,125}{2}.3=0,1875mol\)      \(\Rightarrow V_{H_2}=0,1875.22,4=4,2l\)

\(n_{AlCl_3}=n_{Al}=0,125mol\)       \(\Rightarrow m_{AlCl_3}=0,125.133,5=16,6875g\)