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a) \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
\(n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)
PTHH : Zn + 2HCl -> ZnCl2 + H2
0,25 0,5 0,5 0,5
Xét tỉ lệ : \(\dfrac{0,3}{1}>\dfrac{0,5}{2}\) => Zn dư , HCl đủ
b) \(m_{Zn\left(dư\right)}=\left(0,3-0,25\right).65=3,25\left(g\right)\)
c) \(m_{ZnCl_2}=0,25.136=34\left(g\right)\)
\(V_{H_2}=0,25.22,4=5,6\left(l\right)\)
\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\\ n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\\ a,Zn+2HCl\rightarrow ZnCl_2+H_2\\b, Vì:\dfrac{0,5}{2}< \dfrac{0,3}{1}\Rightarrow Zndư\\ n_{Zn\left(dư\right)}=0,3-\dfrac{0,5}{2}=0,05\left(mol\right)\\ \Rightarrow m_{Zn\left(dư\right)}=0,05.65=3,25\left(g\right)\\ c,n_{ZnCl_2}=n_{H_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ \Rightarrow m_{ZnCl_2}=0,25.136=34\left(g\right)\\ V_{H_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\)
\(a,n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ \text{Vì }\dfrac{n_{Zn}}{1}< \dfrac{n_{HCl}}{2}\text{ nên sau p/ứ }HCl\text{ dư}\\ \Rightarrow n_{H_2}=n_{Zn}=0,2\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,2\cdot22,4=4,48\left(l\right)\\ b,\text{Chất còn dư là }HCl\\ n_{HCl\left(dư\right)}=n_{HCl\text{ đề}}-n_{HCl\text{ phản ứng}}=0,5-0,4=0,1\left(mol\right)\\ \Rightarrow m_{HCl\text{ dư}}=0,1\cdot36,5=3,65\left(g\right)\)
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(m_{HCl}=100.14,6\%=14,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\), ta được HCl dư.
Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b, Theo PT: \(\left\{{}\begin{matrix}n_{HCl\left(pư\right)}=2n_{Zn}=0,2\left(mol\right)\\n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{HCl\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\)
Ta có: m dd sau pư = 6,5 + 100 - 0,1.2 = 106,3 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,1.136}{106,3}.100\%\approx12,79\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,2.36,5}{106,3}.100\%\approx6,87\%\end{matrix}\right.\)
1.
a, \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right);n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)
PTHH: Mg + 2HCl → MgCl2 + H2
Mol: 0,15 0,3
b, Ta có: \(\dfrac{0,15}{1}< \dfrac{0,5}{2}\) ⇒ Mg pứ hết, HCl dư
\(m_{HCldư}=\left(0,5-0,3\right).36,5=7,3\left(g\right)\)
c, \(V_{H_2}=0,15.22,4=3,36\left(l\right)\)
2.
a, \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
PTHH: 4P + 5O2 ---to→ 2P2O5
Mol: 0,2 0,25 0,1
b, \(V_{O_2}=0,25.22,4=5,6\left(l\right)\)
c, \(m_{P_2O_5}=0,1.142=14,2\left(g\right)\)
d, \(V_{kk}=5,6.5=28\left(l\right)\)
nZn=\(\dfrac{13}{65}=0,2\left(mol\right)\)
nHCl= 2 (mol) (theo đề bài)
a) PTHH:
Zn + 2HCl ----> ZnCl2 + H2\(\uparrow\)
0,2.........2..................................................(mol)
b)Tỉ lệ:\(\dfrac{0,2}{1}< \dfrac{2}{2}\) \(\rightarrow\) HCl dư (Tính theo nZn= 0,2 mol)
Theo PTHH: nHCl (pứ) =\(\dfrac{0,2.2}{1}\) =0,4 (mol)
\(\rightarrow\) nHCl (dư) = 2 - 0,4 = 1,6 (mol)
\(\rightarrow\) mHCl (dư) = 1,6 . 36,5= 58,4(gam)
c) Theo PTHH: n\(H_2\) = \(\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(\rightarrow\) V\(H_2\)= 0,2 . 22,4 = 4,48 (l)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\n_{HCl}=0,25\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,25}{2}\) \(\Rightarrow\) HCl còn dư, Kẽm p/ứ hết
\(\Rightarrow\left\{{}\begin{matrix}n_{H_2}=0,1\left(mol\right)\\n_{HCl\left(dư\right)}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=22,4\cdot0,1=2,24\left(l\right)\\m_{HCl\left(dư\right)}=0,05\cdot36,5=1,825\left(g\right)\end{matrix}\right.\)
\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\\ LTL:\dfrac{0,2}{1}< \dfrac{0,5}{2}\Rightarrow HCldư\\ n_{HCl\left(pứ\right)}=2n_{Zn}=0,4\left(mol\right)\\\Rightarrow m_{HCl\left(dư\right)}=\left(0,5-0,4\right).36,5=3,65\left(g\right)\\ b.n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\ \Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\\ c.n_{H_2}=n_{Zn}=0,2\left(mol\right)\\ \Rightarrow V_{H_2}=0,2.22,4,=4,48\left(l\right)\\ d.3H_2+Fe_2O_3-^{t^o}\rightarrow2Fe+3H_2O \\ n_{Fe_2O_3}=\dfrac{19,2}{160}=0,12\left(mol\right)\\ LTL:\dfrac{0,2}{3}< \dfrac{0,12}{1}\Rightarrow Fe_2O_3dưsauphảnứng\\ \Rightarrow n_{Fe}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{15}\left(mol\right)\\ \Rightarrow m_{Fe}=\dfrac{2}{15}.56=7,467\left(g\right)\)
a) n\(Zn\)=\(\dfrac{m}{M}\)=\(\dfrac{13}{65}\)=0,2(mol)
n\(HCl\)=\(\dfrac{m}{M}\)=\(\dfrac{18,25}{36,5}=\)0,5(mol)
PTHH : Zn + 2HCl->ZnCl\(2\) + H\(2\)
0,2 0,5
Lập tỉ lệ mol : \(^{\dfrac{0,2}{1}}\)<\(\dfrac{0,5}{2}\)
n\(Zn\) hết , n\(HCl\) dư
-->Tính theo số mol hết
Zn + 2HCl->ZnCl\(2\) + H\(2\)
0,2 -> 0,4 0,2 0,2
n\(HCl\) dư= n\(HCl\)(đề) - n\(HCl\)(pt)= 0,5 - 0,4 = 0,1(mol)
m\(HCl\) dư= 0,1.36,5 = 3,65(g)
b) m\(ZnCl2\) = n.M= 0,2.136= 27,2 (g)
c)V\(H2\)=n.22,4=0,2.22,4=4,48(l)
d) n\(Fe\)\(2\)O\(3\)=\(\dfrac{m}{M}\)=\(\dfrac{19,2}{160}\)=0,12 (mol)
3H2 +Fe2O3 → 2Fe + 3H2O
0,2 0,12
Lập tỉ lệ mol: \(\dfrac{0,2}{3}\)<\(\dfrac{0,12}{1}\)
nH2 hết .Tính theo số mol hết
\(HCl\)
3H2 +Fe2O3 → 2Fe + 3H2O
0,2-> 0,2
m\(Fe\)=n.M= 0,2.56= 11,2(g)
a) PTHH: \(BaCO_3+2HCl\rightarrow BaCl_2+H_2O+CO_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{BaCO_3}=\dfrac{68,95}{197}=0,35\left(mol\right)\\n_{HCl}=0,25\cdot3,2=0,8\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,35}{1}< \dfrac{0,8}{2}\) \(\Rightarrow\) Axit còn dư
\(\Rightarrow n_{HCl\left(dư\right)}=0,8-0,35\cdot2=0,1\left(mol\right)\) \(\Rightarrow m_{HCl\left(dư\right)}=0,1\cdot36,5=3,65\left(g\right)\)
b+c) Theo PTHH: \(\left\{{}\begin{matrix}n_{CO_2}=n_{BaCl_2}=0,35\left(mol\right)\\n_{HCl\left(dư\right)}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{CO_2}=0,35\cdot22,4=7,84\left(l\right)\\C_{M_{BaCl_2}}=\dfrac{0,35}{0,25}=1,4\left(M\right)\\C_{M_{HCl\left(dư\right)}}=\dfrac{0,1}{0,25}=0,4\left(M\right)\end{matrix}\right.\)
\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
300ml = 0,3l
\(n_{HCl}=2.0,3=0,6\left(mol\right)\)
Pt : \(2Al+6HCl\rightarrow2AlCl_3+3H_2|\)
2 6 2 3
0,3 0,6
Lập tỉ số so sánh : \(\dfrac{0,3}{2}>\dfrac{0,6}{6}\)
⇒ Al dư , HCl phản ứng hết
⇒ Tính toán dựa vào số mol của HCl
\(n_{Al\left(dư\right)}=0,3-\left(\dfrac{0,6.2}{6}\right)=0,1\left(mol\right)\)
⇒ \(m_{Al\left(dư\right)}=0,1.27=2,7\left(g\right)\)
Chúc bạn học tốt
\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right);n_{HCl}=2.0,3=0,6\left(mol\right)\)
PTHH: 2Al + 6HCl → 2AlCl3 + 3H2
Mol: 0,2 0,6
Ta có: \(\dfrac{0,3}{2}>\dfrac{0,6}{6}\) ⇒ Al dư, HCl hết
\(m_{Aldư}=\left(0,3-0,2\right).27=2,7\left(g\right)\)
nZn=19,5/65=0,3(mol)
mHCl=18,25/36,5=0,5(mol)
pt: Zn+2HCl--->ZnCl2+H2
1______2
0,3_____0,5
Ta có: 0,3/1>0,5/2
=>Zn dư
mZn dư=0,05.65=3,25(mol)
Theo pt: nH2=1/2nHCl=1/2.0,5=0,25(mol)
=>VH2=0,25.22,4=5,6(l)
nZn = 0,3 mol
nHCl = 0,5 mol
Zn + 2HCl → ZnCl2 + H2
Đặt tỉ lệ ta có
0,3 < \(\dfrac{0,52}{2}\)
⇒ Zn dư và dư 3,25 gam
⇒ VH2 = 0,25.22,4 = 5,6 (l)