Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(m_{HCl}=100.14,6\%=14,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\), ta được HCl dư.
Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b, Theo PT: \(\left\{{}\begin{matrix}n_{HCl\left(pư\right)}=2n_{Zn}=0,2\left(mol\right)\\n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{HCl\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\)
Ta có: m dd sau pư = 6,5 + 100 - 0,1.2 = 106,3 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,1.136}{106,3}.100\%\approx12,79\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,2.36,5}{106,3}.100\%\approx6,87\%\end{matrix}\right.\)
\(n_{FeO}=\dfrac{10,8}{72}=0,15\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{100.20\%}{100\%}:98=\dfrac{10}{49}\left(mol\right)\)
\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)
0,15----> 0,15 -----> 0,15----->0,15
Xét \(\dfrac{10}{49}:1>\dfrac{0,15}{1}\) => axit dư.
Sau khi phản ứng kết thúc, trong dung dịch có:
\(m_{H_2SO_4}=\left(\dfrac{10}{49}-0,15\right).98=5,3\left(g\right)\)
\(m_{FeSO_4}=0,15.152=22,8\left(g\right)\)
\(m_{H_2O}=0,15.18=2,7\left(g\right)\)
a, \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
b, Ta có: \(m_{H_2SO_4}=200.9,8\%=19,6\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
Theo PT: \(n_{MgO}=n_{MgSO_4}=n_{H_2SO_4}=0,2\left(mol\right)\)
\(\Rightarrow m_{MgO}=0,2.40=8\left(g\right)\)
c, Ta có: m dd sau pư = 8 + 200 = 208 (g)
\(\Rightarrow C\%_{MgSO_4}=\dfrac{0,2.120}{208}.100\%\approx11,54\%\)
a, Zn + 2HCl ----> ZnCl2 + H2↑
b, nZn= 6,5:65= 0,1 mol; nHCl= (100*14,6%)/36,5 = 0,4 mol
Zn + 2HCl ----> ZnCl2 + H2↑
trc pư: 0,1 0,4 (mol)
pư: 0,1 0,2 (mol)
sau pư:0 0,2 0,1 0,1 (mol)
VH2(dktc)= 0,1*22,4= 2,24 (L)
c, mZnCl2= 0,1* 136= 13,6 g
mHCl= 0,2* 36,5= 7,3 g
mdd = 6,5 +100 - 0,1*2 =106,3 g
C%ZnCl2= 13,6/106,3* 100%= 12,8%
C%HCl= 7,3/106,3*100%=6,8%
\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)
\(n_{HCl}=\dfrac{100\cdot14.6\%}{36.5}=0.4\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(1........2\)
\(0.1......0.4\)
\(LTL:\dfrac{0.1}{1}< \dfrac{0.4}{2}\Rightarrow HCldư\)
\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)
\(m_{\text{dung dịch sau phản ứng}}=6.5+100-0.1\cdot2=106.3\left(g\right)\)
\(C\%ZnCl_2=\dfrac{0.1\cdot136}{106.3}\cdot100\%=12.79\%\)
\(C\%HCl\left(dư\right)=\dfrac{\left(0.4-0.2\right)\cdot36.5}{106.3}\cdot100\%=6.87\%\%\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\
n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\\
pthh:Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
\(LTL:\dfrac{0,2}{1}< \dfrac{0,25}{1}\)
=> H2SO4 dư
\(n_{H_2}=n_{H_2SO_4\left(p\text{ư}\right)}=n_{Fe}=0,2\left(mol\right)\\
V_{H_2}=0,2.22,4=4,48l\\
m_{H_2SO_4\left(d\right)}=\left(0,25-0,2\right).98=4,9g\)
\(a) Fe + 2HCl \to FeCl_2\\ b) n_{HCl} = \dfrac{182,5.5\%}{36,5} = 0,25(mol)\\ n_{FeCl_2} = n_{H_2} = n_{Fe} = \dfrac{1}{2}n_{HCl} = 0,125(mol)\\ \Rightarrow m_{Fe} = 0,125.56 = 7(gam) ; V = 0,125.22,4 = 2,8(lít)\\ c) m_{dd\ sau\ phản\ ứng} = m_{Fe} + m_{dd\ HCl} - m_{H_2} = 7 + 182,5 - 0,125.2 = 189,25(gam)\\ C\%_{FeCl_2} = \dfrac{0,125.127}{189,25}.100\% = 8,39\%\)
\(\text{1)}m_{KOH}=40.35\%=14\left(g\right)\\ \rightarrow n_{KOH}=\dfrac{14}{56}=0,25\left(mol\right)\\ PTHH:KOH+HCl\rightarrow KCl+H_2O\\ \text{Theo pthh}:n_{HCl}=n_{KOH}=0,25\left(mol\right)\\ \rightarrow V_{ddHCl}=0,25.0,5=0,125\left(l\right)\)
\(\text{2)}n_{Al}=\dfrac{4,05}{27}=0,15\left(mol\right)\\ n_{H_2SO_4}=200.14,7\%=29,4\left(g\right)\\ \rightarrow n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\\ \text{PTHH}:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\\ \text{LTL}:\dfrac{0,15}{2}< \dfrac{0,3}{3}\rightarrow H_2SO_4\text{ dư}\)
\(\text{Theo pthh}:\left\{{}\begin{matrix}n_{H_2SO_4\left(pư\right)}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,15=0,225\left(mol\right)\\n_{H_2}=n_{H_2SO_4\left(pư\right)}=0,225\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}.0,15=0,075\left(mol\right)\end{matrix}\right.\\ \rightarrow m_{dd\left(\text{sau phản ứng}\right)}=200+4,05-0,3.2=203,45\left(g\right)\)
\(\rightarrow\left\{{}\begin{matrix}C\%_{H_2SO_4\text{ dư}}=\dfrac{\left(0,3-0,225\right).98}{203,45}=3,61\%\\C\%_{Al_2\left(SO_4\right)_3}=\dfrac{342.0,075}{203,45}=12,61\%\end{matrix}\right.\)
a)+b)nCuO=1,6:80=0,02(mol)
\(m_{H_2SO_4}\)=19,6%.100=19,6(g)
=>\(n_{H_2SO_4}\)=19,6:98=0,2(mol)
* PTHH:
CuO+H2SO4->CuSO4+H2O
0,02....0,02........0,02...............(mol)
Ta có:\(\dfrac{n_{CuO}}{1}\)<\(\dfrac{n_{H_2SO_4}}{1}\)(0,02<0,2)=>CuO hết;H2SO4 dư.Tính theo CuO
Theo PTHH:\(m_{CuSO_4}\)=0,02.160=3,2(g)
\(n_{H_2SO_4\left(dư\right)}\)=0,2-0,02=0,18(mol)
=>\(m_{H_2SO_4\left(dư\right)}\)=0,18.98=17,64(g)
Ta có:mdd(sau)=1,6+100=101,6(g)
=>C%CuO=\(\dfrac{3,2}{101,6}\).100%=3,15%
=>\(C_{\%H_2SO_4}\)=\(\dfrac{17,64}{101,6}\).100%=17,36%
nCuO = \(\dfrac{1,6}{80}=0,02\left(mol\right)\) ; n\(H_2SO_4\) =\(\dfrac{100.19,6}{100.98}=0,2\left(mol\right)\)
a) PTHH:
CuO + H2SO4 ----> CuSO4 + H2O
0,02.......0,2.......................................(mol)
Tỉ lệ: \(\dfrac{0,02}{1}< \dfrac{0,2}{1}\Rightarrow H_2SO_4dư\) (Tính theo nCuO = 0,02 mol)
b) Các chất có trong dung dịch sau khi phản ứng kết thúc gồm H2SO4 dư và CuSO4.
Ta có: n\(H_2SO_4\left(pứ\right)\) = nCuO = 0,02 (mol)
=> n\(H_2SO_4\left(dư\right)\) = 0,2 - 0,02 = 0,18 (mol)
=> m\(H_2SO_4\left(dư\right)\) = 0,18 . 98 = 17,64 (gam)
và: n\(CuSO_4\) = nCuO = 0,02 (mol)
=> m\(CuSO_4\) = 0,02 . 160 = 3,2 (gam)
mà: mdd sau pứ = 1,6 + 100 = 101,6 (gam)
Vậy: C%\(H_2SO_4\left(dư\right)\) = \(\dfrac{17,64}{101,6}.100\%\simeq17,36\%\)
và: C%\(CuSO_4\) = \(\dfrac{3,2}{101,6}.100\%\simeq3,15\%\)