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a) Gọi \(n_{Mg}=4x\left(mol\right)\Rightarrow n_{Al}=5x\left(mol\right)\)
=> \(24.4x+27.5x=6,93\Leftrightarrow x=0,03mol\)
=> \(n_{Mg}=4.0,03=0,12mol\Rightarrow m_{Mg}=2,88g,mAl=6,93-2,88=4,05g\)
b) pt:
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,12 0,24
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,15 0,45
=> nHCl = 0,24+0,45=0,69 mol
=> VHCl = 0,69:4=0,1725 lít
a) \(\left\{{}\begin{matrix}24.n_{Mg}+27.n_{Al}=6,93\\\dfrac{n_{Mg}}{n_{Al}}=\dfrac{4}{5}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}n_{Mg}=0,12\left(mol\right)\\n_{Al}=0,15\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}m_{Mg}=0,12.24=2,88\left(g\right)\\m_{Al}=0,15.27=4,05\left(g\right)\end{matrix}\right.\)
b)
PTHH: Mg + 2HCl --> MgCl2 + H2
0,12->0,24
2Al + 6HCl --> 2AlCl3 + 3H2
0,15-->0,45
=> nHCl(min) = 0,24 + 0,45 =0,69 (mol)
=> \(V_{dd.HCl\left(min\right)}=\dfrac{0,69}{4}=0,1725\left(l\right)\)
Câu 1:
Gọi số mol NaCl, KCl là a, b (mol)
=> 58,5a + 74,5b = 6,81 (1)
\(n_{AgCl}=\dfrac{14,35}{143,5}=0,1\left(mol\right)\)
Bảo toàn Cl: a + b = 0,1 (2)
(1)(2) => a = 0,04 (mol); b = 0,06 (mol)
\(\left\{{}\begin{matrix}m_{NaCl}=0,04.58,5=2,34\left(g\right)\\m_{KCl}=0,06.74,5=4,47\left(g\right)\end{matrix}\right.\)
Câu 2:
Gọi số mol MgCl2, KCl là a, b (mol)
=> 95a + 74,5b = 3,93 (1)
25ml dd A chứa \(\left\{{}\begin{matrix}MgCl_2:0,05a\left(mol\right)\\KCl:0,05b\left(mol\right)\end{matrix}\right.\)
nAgNO3 = 0,05.0,06 = 0,003 (mol)
=> nAgCl = 0,003 (mol)
Bảo toàn Cl: 0,1a + 0,05b = 0,003 (2)
(1)(2) => a = 0,01 (mol); b = 0,04 (mol)
\(\left\{{}\begin{matrix}\%m_{MgCl_2}=\dfrac{0,01.95}{3,93}.100\%=24,173\%\\\%m_{KCl}=\dfrac{0,04.74,5}{3,93}.100\%=75,827\%\end{matrix}\right.\)
$a)PTHH:2Al+6HCl\to 2AlCl_3+3H_2$
$n_{H_2}=\dfrac{5,04}{22,4}=0,225(mol)$
$\Rightarrow n_{Al}=0,15(mol)$
$\Rightarrow \%m_{Al}=\dfrac{0,15.27}{9,45}.100\%\approx 42,86\%$
$\Rightarrow \%m_{Cu}=100-42,86=57,14\%$
$b)$ Theo PT: $n_{HCl}=2n_{H_2}=0,45(mol)$
$\Rightarrow C_{M_{HCl}}=\dfrac{0,45.110\%}{0,5}=0,99M$
Ta có: \(m_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\) \(\Rightarrow m_{H_2}=0,2\cdot2=0,4\left(g\right)\)
\(\Rightarrow m_{dd\left(sau.pư\right)}=m_{hh}+m_{ddH_2SO_4}-m_{H_2}=309,6\left(g\right)\)
\(\Rightarrow a=309,6-300=9,6\left(g\right)\)
1)
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\Rightarrow n_{HCl}=0,25.2=0,5\left(mol\right)\)
\(V=\dfrac{0,5}{0,5}=1\left(l\right)\)
2)
\(n_{NaCl}=\dfrac{5,85}{58,5}=0,1\left(mol\right)\); \(n_{AgNO_3}=\dfrac{34}{170}=0,2\left(mol\right)\)
PTHH: NaCl + AgNO3 --> NaNO3 + AgCl
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\) => NaCl hết, AgNO3 dư
PTHH: NaCl + AgNO3 --> NaNO3 + AgCl
0,1------------------------>0,1
=> mAgCl = 0,1.143,5 = 14,35 (g)