Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,n_{K_2O}=\dfrac{18,8}{94}=0,2\left(mol\right)\)
PTHH: K2O + H2O ---> 2KOH
0,2--------------->0,4
b, \(V_{ddKOH}=\dfrac{0,4}{2}=0,2\left(l\right)\)
c, dd sau phản ứng là QT chuyển sang màu xanh vì KOH là dd bazơ
Bài 14 :
\(a) n_{CuO} = \dfrac{8}{80} = 0,1(mol)\\ CuO + 2HCl \to CuCl_2 + H_2O\\ n_{HCl} = 2n_{CuO} = 0,2(mol)\\ m_{dd\ HCl} = \dfrac{0,2.36,5}{7,3\%} = 100(gam)\\ b) \text{Chất tan : } CuCl_2\\ n_{CuCl_2} = n_{CuO} = 0,1(mol)\\ m_{CuCl_2} = 0,1.135 = 13,5(gam)\)
Bài 15 :
\(a) n_{Fe_2O_3} =\dfrac{4,8}{160} = 0,03(mol)\\ Fe_2O_3 + 3H_2SO_4 \to Fe_2(SO_4)_3 + 3H_2O\\ n_{H_2SO_4} = 3n_{Fe_2O_3} = 0,09(mol)\\ m_{dd\ H_2SO_4} = \dfrac{0,09.98}{9,8\%} = 90(gam)\\ b) \text{Chất tan : } Fe_2(SO_4)_3\\ n_{Fe_2(SO_4)_3} = n_{Fe_2O_3} = 0,03(mol)\\ m_{Fe_2(SO_4)_3} = 0,03.400 = 12(gam)\)
Bài 1:
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,4\left(mol\right)\\n_{H_2}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,4\cdot36,5}{14,6\%}=100\left(g\right)\\V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\end{matrix}\right.\)
Bài 2:
PTHH: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{KOH}=\dfrac{100\cdot11,2\%}{56}=0,2\left(mol\right)\\n_{H_2SO_4}=\dfrac{150\cdot9,8\%}{98}=0,15\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,15}{1}\) \(\Rightarrow\) H2SO4 còn dư, KOH p/ứ hết
\(\Rightarrow\left\{{}\begin{matrix}n_{K_2SO_4}=0,1\left(mol\right)\\n_{H_2SO_4\left(dư\right)}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{K_2SO_4}=0,1\cdot174=17,4\left(g\right)\\m_{H_2SO_4\left(dư\right)}=0,05\cdot98=4,9\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{ddKOH}+m_{ddH_2SO_4}=250\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{K_2SO_4}=\dfrac{17,4}{250}\cdot100\%=6,96\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{4,9}{250}\cdot100\%=1,96\%\end{matrix}\right.\)
a)
$n_{HCl} = 0,2.2 = 0,4(mol)$
$n_{Zn} = \dfrac{9,75}{65}= 0,15(mol)$
$Zn + 2HCl \to ZnCl_2 + H_2$
Ta thấy :
n Zn / 1 = 0,15 < n HCl / 2 = 0,2 nên HCl dư
n H2 = n Zn = 0,15(mol)
V H2 = 0,15.22,4 = 3,36 lít
b)
n HCl pư = 2n Zn = 0,3(mol)
=> n HCl dư = 0,4 - 0,3 = 0,1(mol)
n ZnCl2 = n Zn = 0,15(mol)
CM HCl = 0,1/0,2 = 0,5M
CM ZnCl2 = 0,15/0,2 = 0,75M
c)
Dung dịch A làm quỳ tím hóa đỏ vì có HCl dư
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Zn}=\dfrac{9,75}{65}=0,15\left(mol\right)\\n_{HCl}=0,2\cdot2=0,4\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,4}{2}\) \(\Rightarrow\) HCl còn dư, Kẽm p/ứ hết
\(\Rightarrow\left\{{}\begin{matrix}n_{ZnCl_2}=n_{H_2}=0,15\left(mol\right)\\n_{HCl\left(dư\right)}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\\C_{M_{ZnCl_2}}=\dfrac{0,15}{0,2}=0,75\left(M\right)\\C_{M_{HCl\left(dư\right)}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\end{matrix}\right.\)
Vì HCl còn dư, nên dd sau p/ứ làm quỳ tím hóa đỏ
mH2SO4= \(\dfrac{300.7,35}{100}=22,05g\)
nH2SO4= \(\dfrac{22,05}{98}=0,225 mol\)
mHCl= \(\dfrac{200.7,3}{100}=14,6g\)
nHCl= \(\dfrac{14,6}{36,5}=0,4mol\)
H2SO4 + 2HCl → 2H2O + Cl2 ↑+ SO2 ↑
n trước pư 0,225 0,4
n pư 0,2 ← 0,4 → 0,4 → 0,2 → 0,2 mol
n sau pư dư 0,025 hết
a) mCl2= 0,2. 71= 14,2g
mSO2= 64. 0,2= 12,8g
mH2O= 18. 0,4=7,2g
mdd sau pư= 300 +200 -14,2 -12,8= 473g
C%dd H2O= \(\dfrac{7,2.100}{473}=1,52\)%
b) Mg + 2H2O → Mg(OH)2 + H2 ↑
x → 2x → x → x
Fe + 2H2O → Fe(OH)2 + H2↑
y → 2y → y → y
Gọi x,y lần lượt là số mol của Mg,Fe.
Ta có hệ phương trình:
24x + 56y = 8,7 x= \(\dfrac{5}{64}\)
⇒
2x + 2y = 0,4 y= \(\dfrac{39}{320}\)
VH2= 22,4. \((\dfrac{5}{64}+\dfrac{39}{320})\)= 4,48l
mhh MG(OH)2, Fe(OH)2= 8,7 +250 - 2.(\(\dfrac{5}{64}+\dfrac{39}{320}\)) = 2258,3g
mMg=24. \(\dfrac{5}{64}\)=1.875g
mFe= 8,7-1,875= 6,825g
a) \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
0,4-->0,6---------->0,2------->0,6
=> \(C_{M\left(dd.H_2SO_4\right)}=\dfrac{0,6}{0,15}=4M\)
b) VH2 = 0,6.22,4 = 13,44 (l)
c) \(C_{M\left(Al_2\left(SO_4\right)_3\right)}=\dfrac{0,2}{0,15}=\dfrac{4}{3}M\)
\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\
pthh:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,4 0,6 0,2 0,6
\(C_M_{H_2SO_4}=\dfrac{0,6}{0,15}=4M\\ V_{H_2}=0,622,4=13,44L\)
\(C_M=\dfrac{0,2}{0,15}=1,3M\)
\(\text{1)}m_{KOH}=40.35\%=14\left(g\right)\\ \rightarrow n_{KOH}=\dfrac{14}{56}=0,25\left(mol\right)\\ PTHH:KOH+HCl\rightarrow KCl+H_2O\\ \text{Theo pthh}:n_{HCl}=n_{KOH}=0,25\left(mol\right)\\ \rightarrow V_{ddHCl}=0,25.0,5=0,125\left(l\right)\)
\(\text{2)}n_{Al}=\dfrac{4,05}{27}=0,15\left(mol\right)\\ n_{H_2SO_4}=200.14,7\%=29,4\left(g\right)\\ \rightarrow n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\\ \text{PTHH}:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\\ \text{LTL}:\dfrac{0,15}{2}< \dfrac{0,3}{3}\rightarrow H_2SO_4\text{ dư}\)
\(\text{Theo pthh}:\left\{{}\begin{matrix}n_{H_2SO_4\left(pư\right)}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,15=0,225\left(mol\right)\\n_{H_2}=n_{H_2SO_4\left(pư\right)}=0,225\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}.0,15=0,075\left(mol\right)\end{matrix}\right.\\ \rightarrow m_{dd\left(\text{sau phản ứng}\right)}=200+4,05-0,3.2=203,45\left(g\right)\)
\(\rightarrow\left\{{}\begin{matrix}C\%_{H_2SO_4\text{ dư}}=\dfrac{\left(0,3-0,225\right).98}{203,45}=3,61\%\\C\%_{Al_2\left(SO_4\right)_3}=\dfrac{342.0,075}{203,45}=12,61\%\end{matrix}\right.\)