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MgCl2 + 2NaOH -> 2NaCl + Mg(OH)2 (1)
Mg(OH)2 -> MgO + H2O (2)
nMgCl2=0,2.0,15=0,03(mol)
nNaOH=0,2.0,2=0,04(mol)
Vì \(\dfrac{0,04}{2}< 0,03\) nên MgCl2 dư 0,1 mol
Theo PTHH 1 ta có:
nMg(OH)2=\(\dfrac{1}{2}\)nNaOH=0,02(mol)
nNaCl=nNaOH=0,04(mol)
Theo PTHH 2 ta có:
nMgO=nMg(OH)2=0,02(mol)
mMgO=40.0,02=0,8(g)
CM dd MgCl2=\(\dfrac{0,01}{0,4}=0,025M\)
CM dd NaCl=\(\dfrac{0,04}{0,4}=0,01M\)
\(n_{CuCl_2}=0,2.0,15=0,03mol\)
CuCl2+2NaOH\(\rightarrow\)Cu(OH)2+2NaCl
0,03......0,06...........0,03........0,06
\(m=m_{Cu\left(OH\right)_2}=0,03.98=2,94gam\)
\(C_{M_{NaCl}}=\dfrac{n}{v}=\dfrac{0,06}{0,2+0,3}=0,12M\)
\(a,PTHH:3NaOH+FeCl_3\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\\ 2Fe\left(OH\right)_3\rightarrow^{t^o}Fe_2O_3+3H_2O\uparrow\\ b,n_{FeCl_3}=1,5\cdot0,2=0,3\left(mol\right)\\ \Rightarrow n_{NaOH}=3n_{FeCl_3}=0,9\left(mol\right)\\ \Rightarrow V_{dd_{NaOH}}=\dfrac{0,9}{2}=0,45\left(l\right)\)
Theo đề: \(\left\{{}\begin{matrix}X:Fe\left(OH\right)_3\\A:NaCl\\Y:Fe_2O_3\end{matrix}\right.\)
Theo PT: \(n_{NaCl}=3n_{FeCl_3}=0,9\left(mol\right)\)
\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,9}{0,45+0,2}\approx1,4M\)
\(c,\) Theo PT: \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=0,3\left(mol\right);n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_X=m_{Fe\left(OH\right)_3}=0,3\cdot107=32,1\left(g\right)\\m_Y=m_{Fe_2O_3}=0,15\cdot160=24\left(g\right)\end{matrix}\right.\)
\(a,PTHH:CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\\ Cu\left(OH\right)_2\rightarrow^{t^0}CuO+H_2O\\ b,n_{CuCl_2}=n_{Cu\left(OH\right)_2}=n_{CuO}=0,2\left(mol\right)\\ \Rightarrow m_{CuO}=0,2\cdot80=16\left(g\right)\\ c,n_{NaCl}=2n_{CuCl_2}=0,4\left(mol\right)\\ \Rightarrow m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\\ \Rightarrow C\%_{NaCl}=\dfrac{23,4}{200}\cdot100\%=11,7\%\)
\(\text{nMgCl2 = 0,2.0,15 = 0,03 mol}\)
\(\text{a) (1) MgCl2 + 2NaOH → Mg(OH)2 ↓ + 2NaCl}\)
\(\text{ (2) Mg(OH)2 --to--> MgO + H2O}\)
b)
Từ các PTHH: nMgO(2) = nMg(OH)2 (2) = nMg(OH)2 (1) = nMgCl2 = 0,03 mol
\(\text{→ m = mMgO = 0,03.40 = 1,2 (g)}\)
c) Theo PTHH (1): nNaCl = 2nMgCl2 = 2.0,03 = 0,06 mol
\(\text{→ CM NaCl = n : V = 0,06 : (0,2 + 0,3) = 0,12M}\)
a) MgCl2+2NaOH---->Mg(OH)2+2NaCl
Mg(OH)2--->MgO+H2O
b) n MgCl2=0,2.0,15=0,03(mol)
Theo pthh1
n Mg(OH)2=n MgCl2=0,03(mol0
Theo pthh2
n MgO=n Mg(OH)2=0,03(mol)
m MgO=0,03.40=1,2(g)
c) V dd =0,2+0,3=0,5(l)
Theo pthh2
n NaCl=2nMgCl2=0,06(mol)
CM NaCl=\(\frac{0,06}{0,5}=0,12\left(M\right)\)