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\(Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{HCl}=0,3.0,15=0,045\left(mol\right)\\ n_{Zn}=n_{H_2}=\dfrac{0,045}{2}=0,0225\left(mol\right)\\ a,m_{Zn}=0,0225.65=1,4625\left(g\right)\\ b,V_{H_2\left(đktc\right)}=22,4.0,0225=0,504\left(l\right)\\ c,n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\\ Vì:\dfrac{0,3}{1}>\dfrac{0,045}{2}\Rightarrow Zndư\\ \Rightarrow n_{ZnCl_2}=\dfrac{0,045}{2}=0,0225\left(mol\right)\\ m_{ZnCl_2}=0,0225.136=3,06\left(g\right)\)
\(a,n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{Mg}=n_{MgCl_2}=n_{H_2}=0,4\left(mol\right)\\ n_{HCl}=2.0,4=0,8\left(mol\right)\\ a,m_{HCl}=0,8.36,5=29,2\left(g\right)\\ b,m_{ddHCl}=\dfrac{29,2}{10\%}=292\left(g\right)\\ m_{ddsau}=0,4.24+292-0,4.2=300,8\left(g\right)\\ C\%_{ddMgCl_2}=\dfrac{0,4.95}{300,8}.100\approx12,633\%\)
Ta có: \(n_{Na_2CO_3}=0,4.1=0,4\left(mol\right)\)
PT: \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)
a, \(n_{HCl}=2n_{Na_2CO_3}=0,8\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,8}{0,6}=\dfrac{4}{3}\left(M\right)\)
b, \(n_{NaCl}=2n_{Na_2CO_3}=0,8\left(mol\right)\Rightarrow m_{NaCl}=0,8.58,5=46,8\left(g\right)\)
\(n_{CO_2}=n_{Na_2CO_3}=0,4\left(mol\right)\Rightarrow V_{CO_2}=0,4.24,79=9,916\left(l\right)\)
c, \(C_{M_{NaCl}}=\dfrac{0,8}{0,4+0,6}=0,8\left(M\right)\)
1) $n_{HCl} = 0,1.2 = 0,2(mol)$
$Na_2CO_3 + 2HCl \to 2NaCl + CO_2 + H_2O$
Theo PTHH :
$n_{Na_2CO_3} = \dfrac{1}{2}n_{HCl} = 0,1(mol)$
$V_{dd\ Na_2CO_3} = \dfrac{0,1}{1} = 0,1(lít) = 100(ml)$
2)
$n_{NaCl} = n_{HCl} = 0,2(mol)$
$m_{NaCl} = 0,2.58,5 = 11,7(gam)$
3)
$n_{CO_2} = n_{Na_2CO_3} = 0,1(mol)$
$V_{CO_2} = 0,1.22,4 = 2,24(lít)$
100ml =0,1l
\(n_{HCl}=2.0,1=0,2\left(mol\right)\)
Pt : \(HCl+Na_2CO_3\rightarrow2NaCl+CO_2+H_2O|\)
1 1 2 1 1
0,2 0,2 0,4 0,2
1) \(n_{Na2CO3}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(V_{ddNa2CO3}=\dfrac{0,2}{1}=0,2\left(l\right)\)
2) \(n_{NaCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
⇒ \(m_{NaCl}=0,4.58,5=23,4\left(g\right)\)
3) \(n_{CO2}=\dfrac{0,4.1}{2}=0,2\left(mol\right)\)
\(V_{CO2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)
Chúc bạn học tốt
1) nZn=13/65=0,2(mol)
PTHH: Zn + 2 HCl -> ZnCl2 + H2
nH2=nZnCl2=nZn=0,2(mol)
nHCl=2.0,2=0,4(mol)
=> mHCl=0,4 x 36,5=14,6(g)
=> mddHCl=(14,6.100)/8=182,5(g)
2) V(H2,đktc)=0,2 x 22,4= 4,48(l)
mZnCl2=0,2.136=27,2(g)
3) mddsau=mZn+mddHCl - mH2= 13+182,5-0,2.2=195,1(g)
4) C%ddZnCl2=(27,2/195,1).100=13,941%
a/ \(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)
PTHH: Mg + 2HCl → MgCl2 + H2
Mol: 0,3 0,6 0,3 0,3
\(m_{Mg}=0,3.24=7,2\left(g\right)\)
b/ \(m_{MgCl_2}=0,3.95=28,5\left(g\right)\)
c/ \(m_{HCl}=0,3.36,5=10,95\left(g\right)\)
a) $ZnO + 2HCl \to ZnCl_2 + H_2O$
Theo PTHH : $n_{ZnCl_2} = n_{ZnO} = \dfrac{32,4}{81} = 0,4(mol)$
$m_{ZnCl_2} = 0,4.136 = 54,4(gam)$
b) $n_{HCl} = 2n_{ZnO} = 0,8(mol)$
$m_{dd\ HCl} = \dfrac{0,8.36,5}{5\%} = 584(gam)$
c) $V_{dd\ HCl} = \dfrac{584}{1,22} = 478,7(ml)$