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\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{H_2}=n_{Mg}=0,2\left(mol\right)\\ n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\\ CuO+H_2\underrightarrow{^{to}}Cu+H_2O\\ Vì:\dfrac{0,25}{1}>\dfrac{0,2}{1}\\ \Rightarrow CuOdư\\ \Rightarrow n_{Cu}=n_{H_2}=0,2\left(mol\right)\\ m_{Cu}=0,2.64=12,8\left(g\right)\)
\(n_{Mg}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)
1 2 1 1
0,2 0,2
\(n_{H2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\)
Pt : \(H_2+CuO\rightarrow\left(t_o\right)Cu+H_2O|\)
1 1 1 1
0,2 0,25 0,2
Lập tỉ số so sánh : \(\dfrac{0,2}{1}< \dfrac{0,25}{1}\)
⇒ H2 phản ứng hết , CuO dư
⇒ Tính toán dựa vào số mol của H2
\(n_{Cu}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{Cu}=0,2.64=12,8\left(g\right)\)
Chúc bạn học tốt
1: \(n_{Zn}=\dfrac{3.25}{65}=0.05\left(mol\right)\)
a: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,05 0,1 0,05 0,05
\(m_{dd\left(HCl\right)}=0.1\cdot36.5=3.65\left(g\right)\)
b: \(V_{H_2}=0.05\cdot22.4=1.12\left(lít\right)\)
2)
H3PO4 (axit yếu) : axit photphoric
Zn3(PO4)2 (muối) : kẽm photphat
Fe2(SO4)3 (muối) : sắt (III) sunfat
SO2 (oxit axit) : lưu huỳnh đioxit
SO3 (oxit axit) : lưu huỳnh trioxit
P2O5 (oxit axit) : đi photpho pentaoxit
HCl(axit mạnh) : axit clohidric
Ca(HCO3)2 (muối axit) : canxi hidrocacbonat
Ca(H2PO4)2 (muối aixt) : canxi đihidrophotphat
Fe2O3 (oxit bazơ) : sắt (III) oxit
Cu(OH)2 (bazơ) : đống(II) hidroxit
NaH2PO4 (muối axit) : natri đihidrophotphat
Chúc bạn học tốt
a, \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
\(m_{HCl}=200.14,6\%=29,2\left(g\right)\Rightarrow n_{HCl}=\dfrac{29,2}{36,5}=0,8\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Xét tỉ lệ: \(\dfrac{0,3}{1}< \dfrac{0,8}{2}\), ta được HCl dư.
Theo PT: \(n_{H_2}=n_{Zn}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
b, \(n_{ZnCl_2}=n_{Zn}=0,3\left(mol\right)\Rightarrow m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)
c, \(n_{HCl\left(pư\right)}=2n_{Zn}=0,6\left(mol\right)\Rightarrow n_{HCl\left(dư\right)}=0,2\left(mol\right)\)
Ta có: m dd sau pư = 19,5 + 200 - 0,3.2 = 218,9 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{0,2.36,5}{218,9}.100\%\approx3,33\%\\C\%_{ZnCl_2}=\dfrac{40,8}{218,9}.100\%\approx18,64\%\end{matrix}\right.\)
\(a)n_{Zn}=\dfrac{19,5}{65}=0,3mol\\ n_{HCl}=\dfrac{200.14,6}{100.36,5}=0,8mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ \Rightarrow\dfrac{0,3}{1}< \dfrac{0,8}{2}\Rightarrow HCl.dư\\ n_{H_2}=n_{ZnCl_2}=n_{Zn}=0,3mol\\ V_{H_2}=0,3.22,4=6,72l\\ b)m_{ZnCl_2}=0,3.136=40,8g\\ c)n_{HCl.pư}=0,3.2=0,6mol\\ C_{\%ZnCl_2}=\dfrac{40,8}{200+19,5-0,3.2}\cdot100=18,64\%\\ C_{\%HCl.dư}=\dfrac{\left(0,8-0,6\right).36,5}{200+19,5-0,3.2}\cdot100=3,33\%\)
200ml = 0,2l
\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,3 0,6 0,3 0,3
a) \(n_{ZnCl2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
\(C_{M_{ZnCl2}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)
b) \(n_{H2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)
c) Pt : \(NaOH+HCl\rightarrow NaCl+H_2O|\)
1 1 1 1
0,6 0,6
\(n_{NaOH}=\dfrac{0,6.1}{1}=0,6\left(mol\right)\)
\(m_{NaOH}=0,6.40=24\left(g\right)\)
\(m_{ddNaOH}=\dfrac{24.100}{20}=120\left(g\right)\)
Chúc bạn học tốt
a) nZn=6,5:65=0,1(mol)
pt: Zn + 2HCl -> ZnCl2 + H2 (1)
theo pt có: nH2=nZn=0,1(mol)
-> VH2=0,1.22,4=2,24(l)
b) Đổi: 100ml=0,1(l)
theo pt (1) có: nHCl=2nH2=2.0,1=0,2(mol)
->CM=nddHCl/VddHCl=0,2:0,1=2(M)
c) pt: H2(k) + CuO(r) -to-> Cu(r) +H2O(l)
theo pt ta có: nH2=nCuO=0,1(mol)
-> mCuO=0,1.80=8(g)
nZn= 19,5/65=0,3(mol); nFe2O3=19,2/160=0,12(mol)
PTHH: Zn + 2 HCl -> ZnCl2 + H2
Fe2O3 + 3 H2 -to-> 2 Fe +3 H2O
nH2=nZnCl2= nZn=0,3(mol) => V(H2,đktc)=0,3.22,4= 6,72(l)
b) nHCl= 2.0,3=0,6(mol) => mHCl=0,6.36,5=21,9(g)
=>mddHCl=(21,9.100)/20=109,5(g)
=>m=109,5(g)
c) mH2=0,3.2=0,6(mol)
mddZnCl2=19,5+109,5 - 0,6= 128,4(g)
mZnCl2=0,3. 136= 40,8(g)
=>C%ddZnCl2= (40,8/128,4).100=31,776%
d) Ta có: 0,3/3 < 0,12/1
=> H2 hết, Fe2O3 dư, tính theo nH2
=> nFe= 2/3. nH2= 2/3. 0,3= 0,2(mol)
=>mFe=0,2.56=11,2(g)
a, nZn = 19,5/65=0,3 (mol)
PTHH: Zn + 2HCl → ZnCl2 + H2
Mol: 0,3 0,15 0,3 0,3
=> \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
b,mHCl=0,15.36,5=5,475 (g)
=> m=mddHCl=5,475:20%=27,375 (g)
c,mdd sau pứ =19,5+27,375=46,875 (g)
\(m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)
\(\Rightarrow C\%_{ZnCl_2}=\dfrac{40,8}{46,875}.100\%=87,04\%\)
d,\(n_{Fe_2O_3}=\dfrac{19,2}{160}=0,12\left(mol\right)\)
PTHH: Fe2O3 + 3H2 → 2Fe + 3H2O
Mol: 0,3 0,2
Tỉ lệ: 0,12/1>0,3/3 ⇒ Fe2O3 dư,H2 pứ hết
=> mFe=0,2.56=11,2 (g)
a) nAl=2,7/27=0,1(mol)
nHCl=14,6/36,5= 0,4(mol)
PTHH: 2Al +6 HCl -> 2 AlCl3 +3 H2
Ta có: 0,1/2 < 0,6/4
=> HCl dư, Al hết, tính theo nAl
=> nAlCl3=nAl=0,1(mol)
=> mAlCl3=0,1.133,5=13,35(g)
b) nH2= 3/2. nAl=3/2. 0,1=0,15(mol)
=>V(H2,đktc)=0,15.22,4=3,36(l)
c) mFe2O3(nguyên chất)= 80%. 38,4=30,72(g)
=>nFe2O3= 30,72/160=0,192(mol)
PTHH: Fe2O3 + 3 H2 -to->2 Fe +3 H2O
Ta có: 0,192/1 > 0,15/3
=> H2 hết, Fe2O3 dư, tính theo nH2
=> nFe= 2/3. nH2= 2/3. 0,15=0,1(mol)
=>mFe=0,1.56=5,6(g)
a,\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right);n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
PTHH: 2Al + 6HCl → 2AlCl3 + 3H2
Mol: 0,1 0,1 0,15
Tỉ lệ:\(\dfrac{0,1}{2}< \dfrac{0,4}{6}\) ⇒ Al pứ hết,HCl dư
\(\Rightarrow m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)
b,\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)
c,\(m_{Fe_2O_3\left(tinhkhiét\right)}=38,4.\left(100\%-20\%\right)=30,72\left(g\right)\)
⇒\(n_{Fe_2O_3}=\dfrac{30,72}{160}=0,192\left(mol\right)\)
PTHH: Fe2O3 + 3H2 → 2Fe + 3H2O
Mol : 0,15 0,1
Tỉ lệ:\(\dfrac{0,192}{1}>\dfrac{0,15}{3}\)⇒ Fe2O3 dư,H2 hết
=> mFe = 0,1.56 =5,6 (g)
a)
$Fe_2O_3 + 3CO \xrightarrow{t^o} 2Fe +3 CO_2$
$Fe + 2HCl \to FeCl_2 + H_2$
$RO + H_2 \xrightarrow{t^o} R + H_2O$
b)
Coi m = 160(gam)$
Suy ra: $n_{Fe_2O_3} = 1(mol)$
Theo PTHH :
$n_{RO} = n_{H_2} = n_{Fe} = 2n_{Fe_2O_3} = 2(mol)$
$M_{RO} = R + 16 = \dfrac{160}{2} = 80 \Rightarrow R = 64(Cu)$
Vậy oxit là CuO
a) Zn + 2HCl → ZnCl2 + H2
nZn = 9,75 : 65 = 0,15 mol
Theo ptpư
nH2 = nZn = 0,15 mol
VH2 = 0,15 . 22,4 = 3,36 lit
b) CuO + H2 →H2O + Cu
nCuO = 20 : 80 = 0,25 mol
nCuO p/ư = nH2 = 0,15 mol
=> Dư CuO
nCu thu được= nH2 = 0,15 mol
mCu= 0,15 x 64 = 9,6 gam
\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,4--->0,8------>0,4------>0,4 (mol)
\(V_{H_2}=0,4.22,4=8,96\left(l\right)\)
\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
0,4-->0,4 (mol)
=> \(m_{Cu}=0,4.64=25,6\left(g\right)\)