Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(CaO+H_2O->Ca\left(OH\right)_2\)
0,2..........................0,2
n CaO = \(\dfrac{11,2}{40+16}=0,2mol\)
m Ca(OH)2 = \(0,2.\left(40+16.2+1.2\right)=14,8g\)
C% Ca(OH)2 = \(\dfrac{14,8}{500}.100=2,96\%\)
a, \(C\%_{KCl}=\dfrac{20}{20+60}.100\%=25\%\)
b, \(C\%=\dfrac{40}{40+150}.100\%\approx21,05\%\)
c, \(C\%_{NaOH}=\dfrac{60}{60+240}.100\%=20\%\)
d, \(C\%_{NaNO_3}=\dfrac{30}{30+90}.100\%=25\%\)
e, \(m_{NaCl}=150.60\%=90\left(g\right)\)
f, \(m_{ddA}=\dfrac{25}{10\%}=250\left(g\right)\)
g, \(n_{NaOH}=120.20\%=24\left(g\right)\)
Gọi: nNaOH (thêm vào) = a (g)
\(\Rightarrow\dfrac{a+24}{a+120}.100\%=25\%\Rightarrow a=8\left(g\right)\)
\(n_{Al}=\dfrac{0,54}{27}=0,02\left(mol\right)\\ m_{H_2SO_4}=9,8\%.40=3,92\left(g\right)\\ n_{H_2SO_4}=\dfrac{3,92}{98}=0,04\left(mol\right)\)
PTHH: 2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2
LTL: \(\dfrac{0,02}{2}< \dfrac{0,04}{3}\rightarrow\)H2SO4 dư
Theo pt: \(\left\{{}\begin{matrix}n_{H_2SO_4\left(pư\right)}=n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,02=0,03\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}.0,02=0,01\left(mol\right)\end{matrix}\right.\)
\(\rightarrow V_{H_2}=0,03.22,4=0,672\left(l\right)\\ m_{dd}=0,54+40=40,54\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}C\%_{Al_2\left(SO_4\right)_3}=\dfrac{342.0,01}{40,54}=8,43\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{\left(0,04-0,03\right).98}{40,54}=2,41\%\end{matrix}\right.\)
\(n_{CO_2}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{NaOH}=\dfrac{m}{M}=\dfrac{m_{dd}.C\%}{M}=\dfrac{200.16\%}{40}=0,8\left(mol\right)\)
Có: \(\dfrac{n_{NaOH}}{n_{CO_2}}=4\)
=> Phản ứng tạo muối Na2CO3
PT:
\(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)
0,2. 0,4 0,2
=> dd sau phản ứng có những chất tan là:
\(\left\{{}\begin{matrix}Na_2CO_3:0,2\left(mol\right)\\NaOH:0,4\left(mol\right)\end{matrix}\right.\)
mdd spu=0,2.44+200=208,8(g)
\(\%m_{NaOH}=\dfrac{0,4.40}{208,8}.100\%=7,66\%\\\%m_{Na_2CO_3}=\dfrac{0,2.106}{208,8}.100\%=10,15\% \)
Câu 1:
nAl= 0,1(mol)
PTHH: 2Al + 6 HCl -> 2 AlCl3 + 3 H2
nAlCl3= nAl=0,1(mol)
-> mAlCl3= 133,5 x 0,1= 13,35(g)
mddAlCl3= mAl + mddHCl - mH2 = 2,7 + 200 - 3/2 x 0,1 x 2= 202,4(g)
C%ddAlCl3= (13,35/202,4).100= 6,596%
PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)
\(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)
\(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)
a) Ta có: \(n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{6,2}{62}=0,2\left(mol\right)\)
\(\Rightarrow C\%_{NaOH}=\dfrac{0,2\cdot40}{193,8+6,2}\cdot100\%=4\%\)
b) Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=0,2\left(mol\right)\\n_{CuSO_4}=\dfrac{200\cdot16\%}{160}=0,2\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,2}{1}\) \(\Rightarrow\) CuSO4 còn dư, NaOH p/ứ hết
\(\Rightarrow n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=n_{CuO}=0,1\left(mol\right)\)
\(\Rightarrow m_{CuO}=0,1\cdot80=8\left(g\right)\)
c) PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
Theo PTHH: \(n_{HCl}=2n_{CuO}=0,2\left(mol\right)\) \(\Rightarrow V_{ddHCl}=\dfrac{0,2}{2}=0,1\left(l\right)=100\left(ml\right)\)
a) - Dung dịch A chứa chất tan NaOH
mddNaOH= 200(g)
=> C%ddNaOH= (4/200).100=2%
