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số mol kẽm tham gia phản ứng là:\(n_{Zn}=\frac{m}{M}=\frac{6,5}{65}=0,1\left(mol\right)\)
PTHH:
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1 (mol)
a, thể tích khí hiđro thu được là:\(V_{H_2}=n_{H_2}\times22,4=0,1\times22,4=2,24\left(l\right)\)
b,khối lượng HCl cần dùng là:\(m_{HCl}=n_{HCl}\times M=0,2\times65=13\left(g\right)\)
a) $n_{Mg} = \dfrac{4,8}{24} = 0,2(mol)$
$Mg + 2HCl \to MgCl_2 + H_2$
Theo PTHH :
$n_{HCl} = 2n_{Mg} = 0,4(mol) \Rightarrow m_{HCl} = 0,4.36,5 = 14,6(gam)$
b)
$n_{MgCl_2} = n_{Mg} = 0,2(mol) \Rightarrow m_{MgCl_2} = 0,2.95 = 19(gam)$
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 0,2.......0,4........0,2.........0,2\left(mol\right)\\ a.m_{HCl}=0,4.36,5=14,6\left(g\right)\\ b.m_{MgCl_2}=0,2.95=19\left(g\right)\)
\(n_{H_2} = \dfrac{448}{1000.22,4} = 0,02(mol)\\ \Rightarrow n_{HCl} = 2n_{H_2} = 0,02.2 = 0,04(mol)\\ m_{muối} = m_X + m_{HCl} - m_{H_2} = 1,53 + 0,04.36,5 - 0,02.2 = 2,95\ gam\)
Bài 1:
\(Na_2O+H_2O\rightarrow2NaOH\\ SO_2+H_2O⇌H_2SO_3\\ HCl+KOH\rightarrow KCl+H_2O\\ Na_2O+2HCl\rightarrow2NaCl+H_2O\\ SO_2+2KOH\rightarrow K_2SO_3+H_2O\\ KOH+SO_2\rightarrow KHSO_3\\ CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ SO_2+Na_2O\rightarrow Na_2SO_3\)
Bài 2 nếu mHCl= 7,3(g) thì đúng hơn 7,1(g) nên anh sửa là 7,3 gam nha em.
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Theo ĐLBTKL,ta có:
\(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\\ \Leftrightarrow6,5+7,3=m_{ZnCl_2}+0,2\\ \Leftrightarrow m_{ZnCl_2}=\left(6,5+7,3\right)-0,2=13,6\left(g\right)\)
Bài 2:
\(a.n_{CaCl_2}=\dfrac{22,2}{111}=0,2\left(mol\right)\\ n_{AgNO_3}=\dfrac{1,7}{170}=0,01\left(mol\right)\\ CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl\\ a.Vì:\dfrac{0,2}{1}>\dfrac{0,01}{2}\Rightarrow CaCl_2dư\\b.n_{AgCl}=n_{AgNO_3}=0,01\left(mol\right)\\ \Rightarrow m_{\downarrow}=m_{AgCl}=143,5.0,01=1,435\left(g\right)\)
Bài 1:
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{HCl}=0,1.3=0,3\left(mol\right)\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ a,Vì:\dfrac{0,3}{2}< \dfrac{0,2}{1}\Rightarrow Zndư\\ b.n_{ZnCl_2}=\dfrac{0,3}{2}=0,15\left(mol\right)\\ m_{ZnCl_2}=136.0,15=20,4\left(g\right)\)
\(2Al+6HCl->2AlCl_3+3H_2\\ Fe+2HCl->FeCl_2+H_2\\ n_{Al}=a;n_{Fe}=b\\ 27a+56b=8,3\\ 1,5a+b=\dfrac{5,6}{22,4}=0,25\\ a=b=0,1\\ m_{Al}=27\cdot0,1=2,7g\\ m_{Fe}=8,3-2,7=5,6g\\ a=\dfrac{3a+2b}{500}\cdot36,5=3,65\%\\ m_{ddsau}=508,3-0,25\cdot2=507,8g\\ C\%_{AlCl_3}=\dfrac{133,5a}{507,8}=2,63\%\\ C\%_{FeCl_2}=\dfrac{127b}{507,8}=2,50\%\)
Đáp án D.
Chất rắn không tan là Cu.
Zn + 2HCl → ZnCl2 + H2
0,2 ← 0,2 (mol)
mZn = 0,2.65 = 13 (g) => mCu = 15 – 13 = 2 (g)
Gọi \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{1,33}{22,4}=0,059375\left(mol\right)\)
PTHH:
Zn + 2HCl ---> ZnCl2 + H2
a a
Fe + 2HCl ---> FeCl2 + H2
b b
Hệ pt \(\left\{{}\begin{matrix}65a+56b=3,73\\a+b=0,059375\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,045\left(mol\right)\\b=0,014375\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{Zn}=0,045.65=2.925\left(g\right)\\m_{Fe}=0,014375.56=0,805\left(g\right)\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{2,925}{3,73}=78,42\%\\\%m_{Fe}=100\%-78,42\%=21,58\%\end{matrix}\right.\)
a) Zn+2HCl---->ZnCl2+H2
b) n H2=4,48/22,4=0,2(mol)
n Zn=n ZnCl2=n H2=0,2(mol)
m Zn=0,2.65=13(g)
m Zncl2=0,2.136=27,2(g)
c) CuO+H2---->Cu+H2O
n CuO=24/80=0,3(mol)
--->CuO dư
n CuO=n H2=0,2(mol)
n CuO dư=0,3-0,2=0,1(g)
m CuO dư=0,1.80=8(g)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2____________0,2____0,2
\(n_{H2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)
\(m_{Zn}=0,2.65=13\left(g\right)\)
\(m_{ZnCl2}=0,2.\left(65+71\right)=27,2\left(g\right)\)
\(PTHH:CuO+H_2\rightarrow Cu+H_2O\)
Ban đầu :0,3____0,2____________
Phứng: 0,2______0,2__________
Sau phứng :0,1___0___________
\(n_{CuO}=\frac{24}{80}=0,3\left(mol\right)\)
Vậy CuO dư
\(m_{CuO_{du}}=0,1.80=8\left(g\right)\)