Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)
$Cu + 2H_2SO_4 \to CuSO_4 + SO_2 + 2H_2O$
$n_{Cu} = n_{SO_2} = \dfrac{1,12}{22,4} = 0,05(mol)$
$\%m_{Cu} = \dfrac{0,05.64}{10}.100\% = 32\%$
$\%m_{CuO} = 100\% -32\% = 68\%$
b)
$NaOH + SO_2 \to NaHSO_3$
$n_{NaOH} = n_{SO_2} = 0,05(mol)$
$V_{dd\ NaOH} = \dfrac{0,05}{2} = 0,025(lít) = 25(ml)$
PT: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
\(Cu+2H_2SO_{4\left(đ\right)}\underrightarrow{t^o}CuSO_4+SO_2+2H_2O\)
Ta có: \(n_{SO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Theo PT: \(n_{Cu}=n_{SO_2}=0,25\left(mol\right)\)
\(\Rightarrow\%m_{Cu}=\dfrac{0,25.64}{24}.100\%\approx66,67\%\)
a. PTHH:
\(Cu+H_2SO_4--\times-->\)
\(CuO+H_2SO_4--->CuSO_4+H_2O\left(1\right)\)
\(Cu+2H_2SO_{4_{đặc}}\overset{t^o}{--->}CuSO_4+SO_2+2H_2O\left(2\right)\)
Ta có: \(n_{SO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
Theo PT(2): \(n_{Cu}=n_{SO_2}=0,05\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,05.64=3,2\left(g\right)\)
\(\Rightarrow\%_{m_{Cu}}=\dfrac{3,2}{10}.100\%=32\%\)
\(\%_{m_{CuO}}=100\%-32\%=68\%\)
\(CuO+H_2SO_{4\left(24,5\%\right)}\rightarrow CuSO_4+H_2O\)
\(Cu+2H_2SO_{4đ}\underrightarrow{t^o}CuSO_4+SO_2+2H_2O\)
\(n_{SO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
\(\Rightarrow n_{Cu}=0,05\left(mol\right)\)
\(\Rightarrow m_{CuO}=10-64.0,05=6,8\left(g\right)\)
\(\Rightarrow n_{CuO}=0,085\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4\left(24,5\%\right)}=0,085\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4\left(24,5\%\right)}=8,33\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4\left(24,5\%\right)}=34\left(g\right)\)
a)
H2SO4(loãng, dư)+CuO→ H2O+ CuSO4(1)
(mol)
H2SO4(loãng, dư)+Cu→không phản ứng
Cu+ 2H2SO4(đặc, nóng)→ CuSO4+ SO2+ 2H2O(2)
(mol) 0,15 0,3 0,15 0,15
b)
\(n_{SO_2}=\dfrac{V}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(m_{Cu}=n.M=0,15.64=9,6\left(gam\right)\)
→\(m_{CuO}=m_{hh}-m_{Cu}=17,6-9,6=8\left(gam\right)\)
=>\(C\%_{Cu}=\dfrac{9,6}{17,6}.100\%=54,54\%\)
\(C\%_{CuO}=\dfrac{8}{17,6}.100\%=0,45\%\)
a/ \(n_{CO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(MgCO_3+H_2SO_4\rightarrow MgSO_4+H_2O+CO_2\)
\(0,5---0,5----0,5---0,5-0,5\)
\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
\(b---b----b-----b\)
\(\Rightarrow m_{MgCO_3}=0,5.\left(24+12+16.3\right)=42\left(g\right)\)
\(\dfrac{m_{MgCO_3}}{m_{MgO}}=\dfrac{7}{3}\Rightarrow m_{MgO}=42.\dfrac{3}{7}=18\left(g\right)\Rightarrow n_{MgO}=b=0,45\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4}=0,45+0,5=0,95\left(mol\right)\) \(\Rightarrow m_{dd}=\dfrac{0,95.98}{0,05}=1862\left(g\right)\)
a) mCu = 3,2 (g)
=> mFe = 6 - 3,2 = 2,8 (g)
\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,05->0,1--->0,05--->0,05
=> V1 = 0,05.22,4 = 1,12 (l)
\(n_{Cu}=\dfrac{3,2}{64}=0,05\left(mol\right)\)
PTHH: 2Fe + 6H2SO4(đ/n) --> Fe2(SO4)3 + 3SO2 + 6H2O
0,05--------------------------------->0,075
Cu + 2H2SO4 --> CuSO4 + SO2 + 2H2O
0,05------------------------>0,05
=> V2 = (0,075 + 0,05).22,4 = 2,8 (l)
b)
nHCl(dư) = 0,5.2 - 0,1 = 0,9 (mol)
=> \(\left\{{}\begin{matrix}C_{M\left(HCl.dư\right)}=\dfrac{0,9}{0,5}=1,8M\\C_{M\left(FeCl_2\right)}=\dfrac{0,05}{0,5}=0,1M\end{matrix}\right.\)
CuO + H2SO4 --> CuSO4 + H2O
Cu + 2H2SO4(đđ) --> CuSO4 + SO2 + 2H2O
b) Theo ptrinh` 2, ta có:
nCu = nSO2 = V/22,4 = 1,12/22,4 = 0,05 mol
=> mCu = n . M = 0,05 . 64 = 3,2 g
theo đề, ta có:
mCu + mCuO = mhỗnhợp
=> mCuO = 10 - 3,2 = 6,8 g
tên khí A là gì vậy đề bài kêu tính % kl mà bạn