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)

cần gấp đây là môn hóa 8 ạ

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:

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

Tk

2Ca + O₂ -> 2CaO (1)

CaO + 2HCl -> CaCl₂ + H₂O (2)

n_Ca=0,4(mol)

n_HCl=0,5(mol)

Từ 1:

n_CaO=n_Ca=0,4(mol)

Vì 1212n_HCl=0,25(mol)

m_CaCl2=111.0,25=27,75(g)

m_CaO=56.0,15=8,4(g)

\(n_{Ca}=\dfrac{8}{40}=0,2mol\\ 2Ca+O_2\xrightarrow[]{t^0}2CaO\\ n_{CaO}=n_{Ca}=0,2mol\\ n_{HCl}=\dfrac{18,25}{36,5}=0,5mol\\ CaO+2HCl\rightarrow CaCl_2+H_2O\\ \Rightarrow\dfrac{0,2}{1}< \dfrac{0,5}{2}\Rightarrow HCl.dư\\ n_{CaCl_2}=n_{CaO}=0,2mol\\ n_{HCl}=2n_{CaO}=0,4mol\\ m_{CaCl_2}=0,2.111=22,2g\\ m_{HCl.dư}=\left(0,5-0,4\right).36,5=3,65g\)

\(a.n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{17,8\%.200}{36,5}=\dfrac{356}{365}\left(mol\right)\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,1}{1}< \dfrac{\dfrac{356}{365}}{2}\\ \Rightarrow Znhết,HCldư\\ n_{HCl\left(dùng\right)}=0,1.2=0,2\left(mol\right)\\ m_{HCl\left(dùng\right)}=0,2.36,5=7,3\left(g\right)\\ b.n_{H_2}=n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ c.n_{HCl\left(Dư\right)}=\dfrac{356}{365}-0,2=\dfrac{283}{365}\left(mol\right)\\ C\%_{ddZnCl_2}=\dfrac{0,1.136}{6,5+200}.100\approx6,586\%\)

\(C\%_{ddHCl\left(dư\right)}=\dfrac{\dfrac{283}{365}.36,5}{6,5+200}.100\approx13,705\%\)

a, PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Ta có: \(n_{CuO}=\dfrac{3,2}{80}=0,04\left(mol\right)\)

Theo PT: \(n_{Cu}=n_{H_2O}=n_{CuO}=0,04\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Cu}=0,04.64=2,56\left(g\right)\\m_{H_2O}=0,04.18=0,72\left(g\right)\end{matrix}\right.\)

b, PT: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)

Ta có: \(n_{Fe_3O_4}=\dfrac{10,8}{232}=\dfrac{27}{580}\left(mol\right)\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{\dfrac{27}{580}}{1}< \dfrac{0,2}{4}\), ta được H₂ dư.

Theo PT: \(\left\{{}\begin{matrix}n_{H_2\left(pư\right)}=n_{H_2O}=4n_{Fe_3O_4}=\dfrac{27}{145}\left(mol\right)\\n_{Fe}=3n_{Fe_3O_4}=\dfrac{81}{580}\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{H_2\left(dư\right)}=0,2-\dfrac{27}{145}=\dfrac{2}{145}\left(mol\right)\)

\(\Rightarrow m_{H_2\left(dư\right)}=\dfrac{2}{145}.2\approx0,0276\left(g\right)\)

\(m_{H_2O}=\dfrac{27}{145}.18\approx3,35\left(g\right)\)

\(m_{Fe}=\dfrac{81}{580}.56\approx7,82\left(g\right)\)

Bạn tham khảo nhé!