Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)
PTHH: Na2CO3 + 2HCl --> 2NaCl + CO2 + H2O
0,2------------------>0,4---->0,2
mdd sau pư = 200 + 21,2 - 0,2.44 = 212,4(g)
=> \(C\%\left(NaCl\right)=\dfrac{0,4.58,5}{212,4}.100\%=11,017\%\)
$PTHH:Na_2CO_3+2HCl\to 2NaCl+H_2O+CO_2\uparrow$
$n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2(mol)$
Theo PT: $n_{NaCl}=n_{CO_2}=0,2(mol)$
$\Rightarrow m_{NaCl}=0,4.58,5=23,4(g);m_{CO_2}=0,2.44=8,8(g)$
$\Rightarrow C\%_{NaCl}=\dfrac{23,4}{21,2+200-8,8}.100\%\approx 11,01\%$
nH2=0.1(mol)
PTHH:Fe+H2SO4->FeSO4+H2
Fe2O3+3H2SO4->Fe2(SO4)3+3H2O
Theo pthh1:nFe=nH2->nFe=0.1(mol)
mFe=0.1*56=5.6(g)->%Fe=5.6:21.6*100=25.9%
%Fe2O3=100-25.9=74.1%
câu b câu c không liên quan đến đề bài bạn ơi,Chỉ có FeSO4,Fe2(SO4)3,và HCl thôi nhé,không có H2SO4 và MgSO4 đâu
a) \(Zn+2HCl->ZnCl_2+H_2\)
x..........2x.........................x................x
\(2Al+6HCl->2AlCl_3+3H_2\)
y.............3y................................y..............1,5y
b) Gọi số mol của Zn và Al lần lượt là x và y
n H2 = 7,616:22,4=0,34 mol => x+1,5y= 0,34 mol
Ta có m muối = 136x+1335y=34,96 g
=> x=0,1 mol , y=0,16 mol
=> a = 65.0,1+27.0,16=10,82 g
V HCl = \(\frac{2.0,1+3.0,16}{1}=0,68lit\)
a/ Fe + 2HCl \(\rightarrow\) FeCl2 + H2
nH2 = \(\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PTHH: nH2 = nFe = 0,15 (mol) \(\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)
\(\Rightarrow m_{Cu}=11-8,4=2,6\left(g\right)\)
\(\Rightarrow\%m_{Fe}=\dfrac{8,4}{11}.100\%\approx76,4\%\)
\(\Rightarrow\%m_{Cu}=100-76,4\approx23,6\%\)
b/ Theo PTHH ta có: nHCl = 2nFe = 2.0,15 = 0,3 (mol)
\(\Rightarrow V_{ddHCl}=\dfrac{0,3}{2}=0,15\left(M\right)\)
c/ mHCl = 36,5 . 0,3 = 10,95(g)
\(\Rightarrow C\%_{HCl}=\dfrac{m_{HCl}}{m_{ddHCl}}.100\%=\dfrac{10,95}{200}.100\%=5,475\%\)
a, PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,15.56}{11}.100\%\approx76,36\%\\\%m_{Cu}\approx23,64\%\end{matrix}\right.\)
b, Theo PT: \(n_{HCl}=2n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow V_{ddHCl}=\dfrac{0,3}{2}=0,15\left(l\right)\)
c, \(C\%_{HCl}=\dfrac{0,3.36,5}{200}.100\%=5,475\%\)
\(m_{NaOH}=\dfrac{200\cdot8}{100}=16\left(g\right)\Rightarrow n_{NaOH}=\dfrac{16}{40}=0,4mol\)
\(NaOH+HCl\rightarrow NaCl+H_2O\)
0,4 0,4 0,4 0,4
a)\(m_{HCl}=0,4\cdot36,5=14,6\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{14,6}{7,3}\cdot100=200\left(g\right)\)
b)\(m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\)
\(m_{H_2O}=0,4\cdot18=7,2\left(g\right)\)
\(m_{ddsau}=200+200-7,2=392,8\left(g\right)\)
\(\Rightarrow C\%=\dfrac{23,4}{392,8}\cdot100=5,96\%\)
c) \(n_{SO_2}=\dfrac{6,72}{22,4}=0,3mol\)
\(2NaOH+SO_2\rightarrow Na_2SO_4+H_2O\)
0,4 0,3 0,3 0,3
\(m_{Na_2SO_4}=0,3\cdot142=42,6\left(g\right)\)
\(a,n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow n_{HCl}=0,4\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,4\cdot36,5=14,6\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{14,6}{200}\cdot100\%=7,3\%\\ b,n_{H_2}=0,2\left(mol\right)\\ \Rightarrow V_{H_2\left(đkc\right)}=0,2\cdot24,79=4,958\left(l\right)\\ c,m_{H_2}=0,2\cdot2=0,4\left(g\right)\\ n_{FeCl_2}=0,2\left(mol\right)\\ \Rightarrow m_{CT_{FeCl_2}}=0,2\cdot127=25,4\left(g\right)\\ \Rightarrow m_{dd_{FeCl_2}}=11,2+200-0,4=210,8\left(g\right)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{25,4}{210,8}\cdot100\%\approx12,05\%\)