thiếu đề ạ ?

a. PTHH : Fe + HCl -> FeCl₂ + H₂

b) \(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)

\(n_{FeCl_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ m_{FeCl_2}=0,25.127=31,75\left(g\right)\)

c) \(n_{H_2}=\dfrac{0,5}{1}=0,5\left(mol\right)\\ V_{H_2}=0,5.22,4=11,2\left(l\right)\)

Mà thể tích tối đa là 10 l -> quả bóng k chứa được hết lượng H2 thoát ra ngoài

Fe+2HCl->FeCl₂+H₂

0,5------------0,5-----0,5 

=>n _Fe=28\56=0,5 mol

=>m _FeCl2=0,5.127=63,5g

=>V_H2=0,5.22,4=11,2l

bóng 10l ko chứa hết đc nhé thừa 1,2l

a,b, \(n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\)

PTHH: Mg + 2HCl ---> MgCl₂ + H₂

           0,25                     0,25     0,25

\(\rightarrow\left\{{}\begin{matrix}m_{MgCl_2}=0,25.95=23.75\left(g\right)\\V_{H_2}=0,25.22,4=5,6\left(l\right)\end{matrix}\right.\)

c, PTHH: PbO + H₂ --t^o--> Pb + H₂O

LTL: \(0,3>0,25\rightarrow\) PbO dư

\(n_{PbO\left(pư\right)}=n_{Pb}=n_{H_2}=0,25\left(mol\right)\\ \rightarrow m_{chất.rắn}=\left(0,3-0,25\right).233+217.0,25=65,9\left(g\right)\) 

nMg = 6 : 24 = 0,25 (mol) 
pthh : Mg + 2HCl -> MgCl2 + H2
        0,25                0,25        0,25 
=> mMgCl2 = 0,25 . 95 = 23,75 (g) 
=> VH2 = 0,25 . 22,4 = 5,6 (L) 
pthh : PbO + H2 -t--> Pb + H2O 
LTL :     \(\dfrac{0,3}{1}\) >   \(\dfrac{0,25}{1}\)
=> PbO dư 
theo pthh : nPb = nH2 = 0,25  (mol) 
=> mPb = 0,25 . 201 = 50,25 (G)

\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right);n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\\ a,Mg+2HCl\rightarrow MgCl_2+H_2\\ Vì:\dfrac{0,1}{2}< \dfrac{0,1}{1}\\ \Rightarrow Mgdư\\ \Rightarrow n_{Mg\left(p.ứ\right)}=n_{MgCl_2}=n_{H_2}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ n_{Mg\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\\ \Rightarrow m_{Mg\left(dư\right)}=0,05.24=1,2\left(g\right)\\ b,m_{MgCl_2}=95.0,05=4,75\left(g\right)\\ c,V_{H_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\)

a. \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

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

PTHH : Mg + 2HCl -> MgCl₂ + H₂

           0,05          0,1     0,05      0,05

Xét tỉ lệ : \(\dfrac{0,1}{1}>\dfrac{0,1}{2}\) => HCl đủ , Mg dư

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

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

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

Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

a, \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

b, \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,2}=2\left(M\right)\)

c, PT: \(HCl+KOH\rightarrow KCl+H_2O\)

Theo PT: \(n_{KOH}=n_{HCl}=0,4\left(mol\right)\)

\(\Rightarrow m_{ddKOH}=\dfrac{0,4.56}{5,6\%}=400\left(g\right)\)

\(\Rightarrow V_{ddKOH}=\dfrac{400}{1,045}\approx382,78\left(ml\right)\)

\(n_{Fe}=\dfrac{28}{56}=0,5mol\)

a, PTHH : \(Fe + 2HCl-> FECl_2+ H_2↑\)

                 0,5                     0,5            0,5        (mol)

b/. Theo phương trình, ta có:

\(n_{FeCl_2}=n_{Fe}=0,5 mol\)

\(m_{FeCl_2}=0,5.127=63,5g\)

c, Thông cảm không biết làm

\(a,PTHH\left(1\right):Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

\(b,n_{Fe}=\dfrac{m}{M}=\dfrac{28}{56}=0,5\left(mol\right)\\ Theo.PTHH\left(1\right):n_{FeCl_2}=n_{Fe}=0,5\left(mol\right)\\ m_{FeCl_2}=n.M=0,5.91,5=45,75\left(g\right)\)

\(c,PTHH\left(2\right):2Mg+O_2\underrightarrow{t^o}2MgO\\ n_{Mg}=\dfrac{m}{M}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ Theo.PTHH\left(2\right):n_{O_2}=2.n_{Mg}=0,2.2=0,4\left(mol\right)\\ V_{O_2\left(đktc\right)}=n.22,4=0,4.22,4=8,96\left(l\right)\)

(c lỗi đề à có oxi chứ ko có hidro nên mik thay bằng oxi nha)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{FeCl_2}=0,1\left(mol\right)=n_{H_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\\m_{FeCl_2}=0,1\cdot127=12,7\left(g\right)\\C_{M_{FeCl_2}}=\dfrac{0,1}{0,1}=1\left(M\right)\\C_{M_{HCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\end{matrix}\right.\)