Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_{Al} = \dfrac{2,7}{27} = 0,1(mol)\\ 2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\\ n_{H_2} = \dfrac{3}{2}n_{Al} = 0,15(mol)\\ \Rightarrow V_{H_2} = 0,15.22,4 = 3,36(lít)\\ n_{HCl} = \dfrac{8,1}{36,5} = \dfrac{81}{365}(mol)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ \dfrac{n_{Al}}{2} = 0,05 > \dfrac{n_{HCl}}{6} = \dfrac{27}{730} \to Al\ dư\\ n_{Al\ pư} = \dfrac{1}{3}n_{HCl} = \dfrac{27}{365}(mol)\\ \)
\(m_{Al\ dư} = 2,7 - \dfrac{27}{265}.27 = 0,703(gam)\)
a) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\) \(\Rightarrow n_{HCl}=0,8\left(mol\right)\) \(\Rightarrow V_{HCl}=\dfrac{0,8}{2}=0,4\left(l\right)\)
b) PTHH: \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{H_2}=0,4\left(mol\right)\\n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Hidro còn dư, CuO p/ứ hết
\(\Rightarrow n_{Cu}=0,3\left(mol\right)\) \(\Rightarrow m_{Cu}=0,3\cdot64=19,2\left(g\right)\)
a) \(n_{H_2SO_4}=\dfrac{200.10\%}{98}=\dfrac{10}{49}\left(mol\right)\)
PTHH: Zn + H2SO4 --> ZnSO4 + H2
\(\dfrac{10}{49}\)------>\(\dfrac{10}{49}\)--->\(\dfrac{10}{49}\)
=> \(V_{H_2}=\dfrac{10}{49}.22,4=\dfrac{32}{7}\left(l\right)\)
b) \(n_{ZnSO_4}=\dfrac{10}{49}\left(mol\right)\)
Bài 2: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Theo PTHH: \(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\)
Theo PTHH: \(n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow m_{FeCl_2}=127\cdot0,1=12,7\left(g\right)\)
a)
$Zn + H_2SO_4 \to ZnSO_4 + H_2(1)$
$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2(2)$
$Fe + H_2SO_4 \to FeSO_4 + H_2(3)$
b)
Coi m Zn = m Al = m Fe = 100(gam)
\(n_{H_2(1)} = n_{Zn} = \dfrac{100}{65}(mol)\\ n_{H_2(2)} = \dfrac{3}{2}n_{Al} = \dfrac{3}{2}.\dfrac{100}{27} = \dfrac{100}{18}(mol)\\ n_{H_2(3)} = n_{Fe} = \dfrac{100}{56}(mol)\\\)
Ta thấy :
\(n_{H_2(1)} < n_{H_2(3)} < n_{H_2(2)}\) nên dùng kim loại Al cho được nhiều khí hidro nhất.
c) Coi $n_{H_2} = 1(mol)$
n Zn = n H2 = 1(mol) => m Zn = 1.65 = 65(gam)
n Al = 3/2 n H2 = 1,5(mol) => m Al = 1,5.27 = 40,5(gam)
n Fe = n H2 = 1(mol) => m Fe = 1.56 = 56(gam)
Vậy cùng một thể tích hidro thì Al có khối lượng nhỏ nhất
a, PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\) (1)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) (2)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\) (3)
b, Giả sử: mZn = mAl = mFe = a (g)
\(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=\dfrac{a}{65}\left(mol\right)\\n_{Al}=\dfrac{a}{27}\left(mol\right)\\n_{Fe}=\dfrac{a}{56}\left(mol\right)\end{matrix}\right.\)
Theo PT: \(\left\{{}\begin{matrix}n_{H_2\left(1\right)}=n_{Zn}=\dfrac{a}{65}\left(mol\right)\\n_{H_2\left(2\right)}=\dfrac{3}{2}n_{Al}=\dfrac{a}{18}\left(mol\right)\\n_{H_2\left(3\right)}=n_{Fe}=\dfrac{a}{56}\left(mol\right)\end{matrix}\right.\)
⇒ Al cho nhiều khí H2 nhất.
c, Giả sử: nH2 (1) = nH2 (2) = nH2 (3) = b (mol)
Theo PT: \(\left\{{}\begin{matrix}n_{Zn}=n_{H_2\left(1\right)}=b\left(mol\right)\\n_{Al}=\dfrac{2}{3}n_{H_2\left(2\right)}=\dfrac{2}{3}b\left(mol\right)\\n_{Fe}=n_{H_2\left(3\right)}=b\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=65b\left(g\right)\\m_{Al}=\dfrac{2}{3}b.27=18b\left(g\right)\\m_{Fe}=56b\left(g\right)\end{matrix}\right.\)
⇒ Khối lượng Al pư là nhỏ nhất.
a)\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(m\right)\)
\(PTHH:Fe+H_2SO_4\xrightarrow[]{}FeSO_4+H_2\)
tỉ lệ :1 1 1 1
số mol :0,1 0,1 0,1 0,1
\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b)\(m_{H_2SO_4}=0,1.98=9,8\left(g\right)\)
PTHH:Fe+H2SO4→FeSO4+H2
nfe=5,6/56=0,1 mol
VO2=0,1.22,4=2,24 (l)
mH2SO4= 0,1.98=9,8(G)
1.
\(n_{H_2}=\dfrac{0.672}{22.4}=0.03\left(mol\right)\)
\(X+2HCl\rightarrow XCl_2+H_2\)
\(0.03.........................0.03\)
\(M_X=\dfrac{1.2}{0.03}=40\)
\(X:Ca\)
2.
\(CT:XCl_2\)
\(XCl_2+2NaOH\rightarrow X\left(OH\right)_2+2NaCl\)
\(X+71.........................X+34\)
\(47.5.............................29\)
\(29\cdot\left(X+71\right)=47.5\cdot\left(X+34\right)\)
\(\Rightarrow X=24\)
\(X:Mg\)
3.
\(2Cu\left(NO_3\right)_2\underrightarrow{^{^{t^0}}}2CuO+4NO_2+O_2\)
\(0.3..........................................0.15\)
\(n=0.3\)
cho em hỏi khúc
"XCl2+2NaOH→X(OH)2+2NaCl
X+71.........................X+34" thì lm răng tính đc 71 và 34 vậy ạ?