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Bài 1:

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

Ta có: \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

\(n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,1}{2}\) , ta được Mg dư.

Theo PT: \(n_{Mg\left(pư\right)}=n_{MgCl_2}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\)

\(\Rightarrow n_{Mg\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\)

\(\Rightarrow m_{Mg\left(dư\right)}=0,05.24=1,2\left(g\right)\)

\(m_{MgCl_2}=0,05.95=4,75\left(g\right)\)

\(V_{H_2}=0,05.22,4=1,12\left(l\right)\)

Bài 2:

PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{14,7}{98}=0,15\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{2}>\dfrac{0,15}{3}\) , ta được Al dư.

Theo PT: \(\left\{{}\begin{matrix}n_{Al\left(pư\right)}=\dfrac{2}{3}n_{H_2SO_4}=0,1\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2SO_4}=0,05\left(mol\right)\\n_{H_2}=n_{H_2SO_4}=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{Al\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)

\(\Rightarrow m_{Al\left(dư\right)}=0,1.27=2,7\left(g\right)\)

\(m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)

\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)

Bài 3:

PT: \(2M+6HCl\rightarrow2MCl_3+3H_2\)

Ta có: \(n_{H_2}=\dfrac{4,704}{22,4}=0,21\left(mol\right)\)

Theo PT: \(n_M=\dfrac{2}{3}n_{H_2}=0,14\left(mol\right)\)

\(\Rightarrow M_M=\dfrac{3,78}{0,14}=27\left(g/mol\right)\)

Vậy: M là nhôm (Al).

Bài 4:

PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=0,2\left(mol\right)\)

PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

Ta có: \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{4}>\dfrac{0,2}{5}\) , ta được P dư.

Theo PT: \(n_{P_2O_5}=\dfrac{2}{5}n_{O_2}=0,08\left(mol\right)\)

\(\Rightarrow m_{P_2O_5}=0,08.142=11,36\left(g\right)\)

Bạn tham khảo nhé!

bạn từng câu lên sẽ dễ nhìn hơn 

A) nZn=0,1(mol); nS=0,2(mol)

PTHH: Zn + S -to-> ZnS

Ta có: 0,2/1 > 0,1/1

=> Zn hết, S dư, tính theo nZnS

=> nZnS= nS(p.ứ)=nZn=0,1(mol)

=> nS(dư)=0,2-0,1=0,1(mol)

=>mS(dư)=0,1.32=3,2(g)

b) mZnS=0,1.81=8,1(g)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{25,55}{36,5}=0,7\left(mol\right)\\a. 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b.Vì:\dfrac{0,2}{2}< \dfrac{0,7}{6}\\ \Rightarrow HCldư\\ n_{HCl\left(dư\right)}=0,7-\dfrac{6}{2}.0,2=0,1\left(mol\right)\\ n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ m_{H_2}=0,3.2=0,6\left(g\right)\\ m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\\ m_{AlCl_3}=133,5.0,2=26,7\left(g\right)\)

Zn+S->ZnS

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

n Zn=\(\dfrac{13}{65}\)=0,2 mol

n S=\(\dfrac{9,6}{32}\)=0,3 mol

=>S dư 

=>m S=0,1.32=3,2g

=>m ZnS=0,2.97=19,4g

a. \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

\(n_S=\dfrac{9.6}{32}=0,3\left(mol\right)\)

Ta thấy : 0,2 < 0,3 => Zn đủ , S dư

PTHH : Zn + S -> ZnS

            0,2     0,2     0,2

\(m_{S\left(dư\right)}=\left(0,3-0,2\right).32=3,2\left(g\right)\)

b. \(m_{ZnS}=0,2.97=19,4\left(g\right)\)

\(pthh:Zn+S\overset{t^o}{--->}ZnS\)

Ta có: \(\left\{{}\begin{matrix}n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\n_S=\dfrac{9,6}{32}=0,3\left(mol\right)\end{matrix}\right.\)

a. Ta thấy: \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\)

Vậy S dư.

Theo pt: \(n_{S_{PỨ}}=n_{Zn}=0,2\left(mol\right)\)

\(\Rightarrow m_{S_{dư}}=\left(0,3-0,2\right).32=3,2\left(g\right)\)

b. Các chất sau phản ứng: \(\left\{{}\begin{matrix}S_{dư}=3,2\left(g\right)\\ZnS\end{matrix}\right.\)

Áp dụng ĐLBTKL, suy ra:

\(m_{ZnS}=13+0,2.32=19,4\left(g\right)\)

12,395l ở đk nào vậy em?

pthhL

\(3O_2+4Al\underrightarrow{t^o}2Al_2O_3\)

\(nO_2=12,395:24,9=0,5\left(mol\right)\)

\(nAl=\dfrac{10,8}{27}=0,4\left(mol\right)\)

Xét tỉ lê:

\(\dfrac{nO_2}{3}>\dfrac{nAl}{4}\)

=> Oxy dư và dư: \(mO_{2dư}=0,06.32=2g\)

b, tính theo pthh vì nO2 dư nên lấy số mol của Al

=> \(nAl_2O_3=\dfrac{2}{4}.0,4=0,2\left(mol\right)\)

\(\Rightarrow mAl_2O_3=0,2.102=20,4\left(g\right)\)

đktc ạ 

\(n_{Al}=\dfrac{13,5}{27}=0,5\left(mol\right)\\ n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ pthh:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\) 
\(LTL:\dfrac{0,5}{4}< \dfrac{0,4}{3}\) 
=> Oxi dư , Al hết 
\(n_{O_2\left(p\text{ư}\right)}=\dfrac{3}{4}n_{Al}=0.375\left(mol\right)\) 
\(n_{O_2\left(d\right)}=0,4-0,375=0,025\left(mol\right)\)
\(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,25\left(mol\right)\\ m_{Al_2O_3}=0,25.102=25,5g\)

a. PTHH: 2Al(OH)₃ + 3H₂SO₄ ---> Al₂(SO₄)₃ + 6H₂O

b. Ta có: \(n_{Al\left(OH\right)_3}=\dfrac{58,5}{78}=0,75\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)

Ta thấy: \(\dfrac{0,75}{2}>\dfrac{0,5}{3}\)

Vậy \(Al\left(OH\right)_3\) dư.

\(m_{dư}=0,75.78-98.0,5=9,5\left(g\right)\)

c. Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}.n_{H_2SO_4}=\dfrac{1}{3}.0,5=\dfrac{1}{6}\left(mol\right)\)

=> \(m_{Al_2\left(SO_4\right)_3}=\dfrac{1}{6}.342=57\left(g\right)\)

a, \(n_{Al\left(OH\right)_3}=\dfrac{58,5}{78}=0,75\left(mol\right);n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)
PTHH: 2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O

Mol:        \(\dfrac{1}{3}\)               0,5               \(\dfrac{1}{6}\)

b, Ta có: \(\dfrac{0,75}{2}>\dfrac{0,5}{3}\) ⇒ Al(OH)₃ dư, H₂SO₄ hết

⇒ \(m_{Al\left(OH\right)_3}=\left(0,75-\dfrac{1}{3}\right).78=32,5\left(g\right)\)

c, \(m_{Al_2\left(SO_4\right)_3}=\dfrac{1}{6}.342=57\left(g\right)\)

Ta có: \(n_{Al\left(OH\right)_3}=\dfrac{58,5}{78}=0,75\left(mol\right)\)

a. PTHH: 2Al(OH)₃ + 3H₂SO₄ ---> Al₂(SO₄)₃ + 6H₂O

b. Không có chất dư (hoặc có thể bn cho sai 49(g) dung dịch là 49(g) H₂SO₄)

c. Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}.n_{Al\left(OH\right)_3}=\dfrac{1}{2}.0,75=0,375\left(mol\right)\)

=> \(m_{Al_2\left(SO_4\right)_3}=0,375.342=128,25\left(g\right)\)