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a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Gọi: \(\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\) ⇒ 65x + 27y = 17,05 (1)
Ta có: \(n_{H_2}=\dfrac{9,52}{22,4}=0,425\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Zn}+\dfrac{3}{2}n_{Al}=x+\dfrac{3}{2}y=0,425\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,2\left(mol\right)\\n_{Al}=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,2.65=13\left(g\right)\\m_{Al}=0,15.27=4,05\left(g\right)\end{matrix}\right.\)
b, Theo PT: \(\left\{{}\begin{matrix}n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\n_{AlCl_3}=n_{Al}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{ZnCl_2}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\\C_{M_{AlCl_3}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\end{matrix}\right.\)
c, Ta có: m dd HCl = 1,05.500 = 525 (g)
m dd sau pư = mhh + m dd HCl - mH2 = 541,2 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,2.136}{541,2}.100\%\approx5,03\%\\C\%_{AlCl_3}=\dfrac{0,15.133,5}{541,2}.100\%\approx3,7\%\end{matrix}\right.\)
1 ) CAO +H2O => CA(OH)2 (1)
2K + 2H2O => 2KOH + H2(2)
n (H2) =1,12/22,4 =0,05
theo ptpư 2 : n(K) = 2n (h2) =2.0.05=0,1(mol)
=> m (K) =39.0,1=3,9 (g)
% K= 3,9/9,5 .100% =41,05%
%ca =100%-41,05%=58,95%
xo + 2hcl =>xcl2 +h2o
10,4/X+16 15,9/x+71
=> giải ra tìm đc X bằng bao nhiêu thì ra
\(n_{HCl}=0,3.2=0,6\left(mol\right)\\a, 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b,n_{H_2}=\dfrac{3}{6}.0,6=0,3\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ c,n_{Al}=n_{AlCl_3}=\dfrac{2}{6}.0,6=0,2\left(mol\right)\\ m_{Al}=0,2.27=5,4\left(g\right)\\ d,V_{ddAlCl_3}=V_{ddHCl}=0,3\left(l\right)\\ C_{MddHCl}=\dfrac{0,2}{0,3}=\dfrac{2}{3}\left(M\right)\)
a) CaO + H2O $\to$ Ca(OH)2
Ca + 2H2O $\to$ Ca(OH)2 + H2
Ca(OH)2 + 2HCl $\to$ CaCl2 + 2H2O
b) n CaO = a(mol) ; n Ca = b(mol)
=> 56a + 40b = 17,2(1)
n HCl = 0,175.2 = 0,35(mol)
=> n Ca(OH)2 = 1/2 n HCl = 0,35/2 = 0,175(mol)
1 nửa dung dịch Y chứa 0,175 mol Ca(OH)2
Vậy Y chứa 0,175 x 2 = 0,35 mol Ca(OH)2
n Ca(OH)2 = a + b = 0,35(2)
Từ (1)(2) suy ra a = 0,2 ; b = 0,15
%m CaO = 0,2.56/17,2 .100% = 65,12%
%m Ca = 100% - 65,12% = 34,88%
c) n CO2 = 5,6/22,4 = 0,25(mol)
Ca(OH)2 + CO2 → CaCO3 + H2O
0,175.......0,175........0,175..................(mol)
CaCO3 + CO2 + H2O→ Ca(HCO3)2
0,075.......0,075....................................(mol)
=> m CaCO3 = (0,175 - 0,075).100 = 10(gam)
a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Ta có: \(n_{Zn}=n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,2.65=13\left(g\right)\)
Bạn xem lại xem đề cho bao nhiêu gam hỗn hợp nhé, vì mZn đã bằng 13 (g) rồi.
a) Mg + 2HCl -> MgCl2 + H2
Al + 3HCl -> AlCl3 + 3/2H2
b) Gọi a, b lần lượt là số mol Mg, Al.
nH2 = 5,6/22,4 = 0,25 (mol)
Mg + 2HCl -> MgCl2 + H2
a 2a a a
Al + 3HCl -> AlCl3 + 3/2H2
b 3b b 3/2b
Ta có hệ pt:
mhh = 24a + 27b = 5,1 (g)
nH2 = a + 3/2b = 0,25 (mol)
=> a = 0,1 (mol)
b = 0,1 (mol)
200 ml = 0,2 l
nHCl = 2a + 3b = 0,2 + 0,3 = 0,5 (mol)
=> CM ddHCl = 0,5/0,2 = 2,5 (M)
%mMg = 24a/5,1*100% = 2,4/5,1*100% = 47,06%
%mAl = 100%-47,06% = 52,94%
Bài 1:
PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\n_{H_2SO_4}=0,3\cdot2=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Axit còn dư
\(\Rightarrow\left\{{}\begin{matrix}n_{FeSO_4}=0,2\left(mol\right)\\n_{H_2SO_4}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{FeSO_4}}=\dfrac{0,2}{0,3}\approx0,67\left(M\right)\\C_{M_{H_2SO_4}}=\dfrac{0,4}{0,3}=0,75\left(M\right)\end{matrix}\right.\)
Mặt khác: \(n_{H_2}=0,2\left(mol\right)\) \(\Rightarrow V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\)
Bài 2:
PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)
a______2a (mol)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
b______2b (mol)
Ta lập HPT: \(\left\{{}\begin{matrix}40a+80b=14\\2a+2b=0,5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,15\\b=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CuO}=0,1\cdot80=8\left(g\right)\\m_{MgO}=6\left(g\right)\end{matrix}\right.\)