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a) $n_{CaCO_3} = 0,15(mol)$
$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
$n_{HCl} = 2n_{CaCO_3} = 0,3(mol)$
$m_{dd\ HCl} = \dfrac{0,3.36,5}{7,3\%} = 150(gam)$
b)
$n_{CaCl_2} = n_{CO_2} = n_{CaCO_3} =0,15(mol)$
$V_{CO_2} = 0,15.22,4 = 3,36(lít)$
c)
$m_{dd} = 15 + 150 - 0,15.44 = 158,4(gam)$
$C\%_{CaCl_2} = \dfrac{0,15.111}{158,4}.100\% = 10,51\%$
\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{HCl}=2\cdot0,15=0,3\left(mol\right)\)
Vì \(\dfrac{n_{Zn}}{1}>\dfrac{n_{HCl}}{2}\) nên sau p/ứ Zn dư
\(\Rightarrow n_{Zn}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\\ \Rightarrow m_{Zn}=0,15\cdot65=9,75\\ \Rightarrow m_{Zn\left(dư\right)}=13-9,75=3,25\left(g\right)\\ c,n_{H_2}=n_{Zn}=0,15\left(mol\right)\\ \Rightarrow V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\)
a)\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{HCL}=0,1.1=0,1\left(mol\right)\)
\(\Rightarrow n_{H_2}=2nH_2=2,0,1=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(ml\right)\)
b) sau pư Fe dư
ta có 1 molFe Pư 2 molHCL
0,05 molFe pư 0,1 HCL
\(\Rightarrow n_{Fe\left(dư\right)}:0,1-0,05=0,05\left(mol\right)\)
c)\(C_{MFeCL_2}=\dfrac{2.n_{HCL}}{0,1}=2M\)
Ta có: nFe=5,656=0,1(mol)
nHCl=0,1.1=0,1(mol)
PT: Fe+2HCl→FeCl2+H2
Xét tỉ lệ: 0,11>0,12, ta được Fe dư.
a, Theo PT: nH2=12nHCl=0,05(mol)
⇒ VH2 = 0,05.22,4 = 1,12 (l)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\n_{HCl}=0,1\cdot1=0,1\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,1}{2}\) \(\Rightarrow\) Sắt còn dư, HCl p/ứ hết
\(\Rightarrow n_{Fe\left(dư\right)}=0,05\left(mol\right)=n_{FeCl_2}\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe\left(dư\right)}=0,05\cdot56=2,8\left(g\right)\\C_{M_{FeCl_2}}=\dfrac{0,05}{0,1}=0,5\left(M\right)\end{matrix}\right.\)
\(n_{Zn}=\dfrac{9,75}{65}=0,15\left(mol\right)\)
\(n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,15 0,4 0,15
a) Lập tỉ số so sánh : \(\dfrac{0,15}{1}< \dfrac{0,4}{2}\)
⇒ Zn phản ứng hết , HCl dư
⇒ Tinsht toán dựa vào số mol của zn
\(n_{HCl\left(dư\right)}=0,4-\left(0,15.2\right)=0,1\left(mol\right)\)
⇒ \(m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\)
b) \(n_{H2}=\dfrac{0,15.1}{1}=01,5\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,15.24,79=3,1875\left(l\right)\)
Chúc bạn học tốt
Ta có: \(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)
\(n_{HCl}=4.100:1000=0,4\left(mol\right)\)
a. PTHH: MgO + 2HCl ---> MgCl2 + H2O
Ta thấy: \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\)
Vậy HCl dư.
=> \(n_{dư}=\dfrac{0,1.2}{0,4}=0,5\left(mol\right)\)
=> \(m_{dư}=0,5.36,5=18,2\left(g\right)\)
b. Ta có: \(V_{dd_{MgCl_2}}=V_{HCl}=\dfrac{100}{1000}=0,1\left(lít\right)\)
Theo PT: \(n_{MgCl_2}=n_{MgO}=0,1\left(mol\right)\)
=> \(C_{M_{MgCl_2}}=\dfrac{0,1}{0,1}=1M\)
\(300(ml)=0,3(l)\\ n_{HCl}=1.0,3=0,3(mol);n_{Fe}=\dfrac{5,6}{56}=0,1(mol)\\ a,PTHH:Fe+2HCl\to FeCl_2+H_2\\ \text{LTL: }\dfrac{n_{Fe}}{1}<\dfrac{n_{HCl}}{2}\Rightarrow HCl\text{ dư}\\ \Rightarrow n_{HCl(dư)}=0,3-0,1.2=0,1(mol)\\ \Rightarrow m_{HCl(dư)}=0,1.36,5=3,65(g)\\ b,n_{FeCl_2}=n_{Fe}=0,1(mol)\\ \Rightarrow \begin{cases} C_{M_{FeCl_2}}=\dfrac{0,1}{0,3}=0,33M\\ C_{M_{HCl(dư)}}=\dfrac{0,1}{0,3}=0,33M \end{cases}\)
\(n_{Fe}=\dfrac{84}{56}=1,5\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=n_{FeCl_2}=n_{Fe}=1,5\left(mol\right)\\ V_{H_2}=1,5.22,4=33,6\left(l\right)\\ C\%_{ddFeCl_2}=\dfrac{127.1,5}{84+300-1,5.2}.100\%=\dfrac{190,5}{381}.100\%=50\%\)
a) nZn=0,15(mol)
pt: Zn +2HCl-> ZnCl2+ H2
=> nH2=0,15(mol)=> V=0,15.22,4=3,36(lít)
b) nHCl=0,4(mol)
sau phản ứng HCl dư: nHCl dư=0,4-0,3=0,1(mol)