Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
PTHH 3ZnCl2+2H3PO4----->Zn3(PO4)2+6HCl
\(n_{ZnCl_2}\)=0,3.2=0,6(mol)
Theo phương trình =>\(\dfrac{1}{3}n_{ZnCl_2}=n_{Zn_3\left(PO_4\right)_2}=0,2\left(mol\right)\)
=>\(m_{Zn_3\left(PO_4\right)_2}\)=0,2.385=77(g)
Theo phương trình =>\(2n_{ZnCl_2}=n_{HCl}=1,2\left(mol\right)\)
=>\(C_{M_{HCl}}\)=\(\dfrac{1,2}{0,2+0,3}=2,4M\)
\(n_{CuCl_2}=0,3.0,5=0,15\left(mol\right)\)
PT: \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\)
Theo PT: \(n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,15\left(mol\right)\Rightarrow m_{Cu\left(OH\right)_2}=0,15.98=14,7\left(g\right)\)
\(n_{NaOH}=2n_{CuCl_2}=0,3\left(mol\right)\)
\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,3}{0,25}=1,2\left(M\right)\)
\(n_{NaCl}=\dfrac{3,51}{58,5}=0,06\left(mol\right)\)
a) Pt : \(NaCl+AgNO_3\rightarrow NaNO_3+AgCl|\)
1 1 1 1
0,06 0,06 0,06 0,06
a) \(n_{AgCl}=\dfrac{0,06.1}{1}=0,06\left(mol\right)\)
⇒ \(m_{AgCl}=0,06.143,5=8,61\left(g\right)\)
b) \(n_{AgNO3}=\dfrac{0,06.1}{1}=0,06\left(mol\right)\)
\(V_{ddAgNO3}=\dfrac{0,06}{0,2}=0,3\left(l\right)\)
c) \(n_{NaNO3}=\dfrac{0,06.1}{1}=0,06\left(mol\right)\)
\(C_{M_{NaNO3}}=\dfrac{0,06}{0,3}=0,2\left(M\right)\)
Chúc bạn học tốt
\(a,n_{NaCl}=\dfrac{3,51}{58,5}=0,06(mol)\\ PTHH:NaCl+AgNO_3\to AgCl\downarrow+NaNO_3\\ \Rightarrow n_{AgCl}=0,06(mol)\\ \Rightarrow m_{AgCl}=0,06.143,5=8,61(g)\\ b,n_{AgNO_3}=0,06(mol)\\ \Rightarrow V_{dd_{AgNO_3}}=\dfrac{0,06}{0,2}=0,3(l)\\ c,n_{NaNO_3}=0,06(mol);V_{dd_{NaNO_3}}=V_{dd(\text {phản ứng})}=0,3(l)\\ \Rightarrow C_{M_{NaNO_3}}=\dfrac{0,06}{0,3}=0,2M\)
a.CuCl2 + 2NaOH -> Cu(OH)2 + 2NaCl
0.15 0.3 0.15 0.3
Cu(OH)2 -> CuO + H2O
0.15 0.15
nNaOH = 0.3 mol
\(CM_{CuCl2}=\dfrac{0.15}{2}=0.075M\)
b.Vdd sau phản ứng = 0.2 + 0.15 = 0.35l
\(CM_{NaCl}=\dfrac{0.3}{0.35}=0.86M\)
c.mCuO = \(0.15\times80=12g\)
\(a,PTHH:CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2+2KCl\\ ...0,2......0,4.......0,2........0,4\left(mol\right)\\ b,n_{CuCl_2}=\dfrac{27}{135}=0,2\left(mol\right)\\ m_{Cu\left(OH\right)_2}=0,2\cdot98=19,6\left(g\right)\\ c,m_{KOH}=0,4\cdot56=22,4\left(g\right)\\ m_{dd_{KOH}}=\dfrac{22,4\cdot100\%}{20\%}=112\left(g\right)\\ m_{dd_{KCl}}=m_{CuCl_2}+m_{dd_{KOH}}-m_{Cu\left(OH\right)_2}=27+112-19,6=119,4\left(g\right)\)
\(d,C\%_{dd_{KCl}}=\dfrac{74,5\cdot0,4}{119,4}\cdot100\%\approx24,96\%\)
HCl + AgNO3 ➜ AgCl↓ + HNO3
\(n_{HCl}=0,2\times2=0,4\left(mol\right)\)
\(n_{AgNO_3}=0,3\times2=0,6\left(mol\right)\)
Theo PT: \(n_{HCl}=n_{AgNO_3}\)
Theo bài: \(n_{HCl}=\dfrac{2}{3}n_{AgNO_3}\)
Vì \(\dfrac{2}{3}< 1\) ⇒ dd HCl hết, dd AgNO3 dư
Theo PT: \(n_{AgCl}=n_{HCl}=0,4\left(mol\right)\)
\(\Rightarrow m_{AgCl}=0,4\times143,5=57,4\left(g\right)\)
Dung dịch B gồm: AgNO3 dư và HNO3
Theo PT: \(n_{AgNO_3}pư=n_{HCl}=0,4\left(mol\right)\)
\(\Rightarrow n_{AgNO_3}dư=0,6-0,4=0,2\left(mol\right)\)
Theo PT: \(n_{HNO_3}=n_{HCl}=0,4\left(mol\right)\)
\(\Sigma n_{ctB}=n_{AgNO_3}dư+n_{HNO_3}=0,2+0,4=0,6\left(mol\right)\)
\(\Sigma m_{ddB}=0,2+0,3=0,5\left(l\right)\)
\(\Rightarrow C_{M_{ddB}}=\dfrac{0,6}{0,5}=1,2\left(M\right)\)
HCl + AgNO3 ➜ AgCl↓ + HNO3
\(n_{HCl}=0,2\times2=0,4\left(mol\right)\)
\(n_{AgNO_3}=0,3\times2=0,6\left(mol\right)\)
Theo PT: \(n_{HCl}=n_{AgNO_3}\)
Theo bài: \(n_{HCl}=\dfrac{2}{3}n_{AgNO_3}\)
Vì \(\dfrac{2}{3}< 1\) ⇒ dd HCl hết, dd AgNO3 dư
Theo PT: \(n_{AgCl}=n_{HCl}=0,4\left(mol\right)\)
\(\Rightarrow m_{AgCl}=0,4\times143,5=57,4\left(g\right)\)
Dung dịch B gồm: AgNO3 dư và HNO3
\(\Sigma m_{ddB}=0,2+0,3=0,5\left(l\right)\)
Theo PT: \(n_{HNO_3}=n_{HCl}=0,4\left(mol\right)\)
Theo PT: \(n_{AgNO_3}pư=n_{HCl}=0,4\left(mol\right)\)
\(\Rightarrow n_{AgNO_3}dư=0,6-0,4=0,2\left(mol\right)\)
\(\Sigma n_{ctB}=n_{AgNO_3}dư+n_{HNO_3}=0,2+0,4=0,6\left(mol\right)\)
\(\Rightarrow C_{M_{ddB}}=\dfrac{0,6}{0,5}=1,2\left(M\right)\)
Giải
nCuCl2 = 0,3 . 0,2 = 0,06 (mol)
nAgNO3 = 0,2 . 0,7 = 0,14 (mol)
\(CuCl_2+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2AgCl\downarrow\)
nTheo bài: 0,06........0,14 mol
nPư: 0,06........0,12 mol
nSau pư: 0.............0,02.............0,06...........0,12 mol
mAgCl = 0,12 . 143,5 = 17,22 (g)
Vdd sau phản ứng= 0,3 + 0,2 = 0,5 (l)
CMCu(NO3)2 = 0,06 : 0,5 = 0,12 M
CMAg(NO3)2 dư = 0,02 : 0,5 = 0,04 M
mdung dịch sau phản ứng= 0,06.135 + 0,14.170 - 0,12.143,5 = 14,68(g)
C%Cu(NO3)2 = \(\dfrac{0,06.188}{14,68}.100\%\simeq76,84\%\)
C%AgNO3 dư = \(\dfrac{0,02.170}{14,68}.100\%\simeq23,16\%\)
Cảm ơn bạn nhiều