Bạn xem lại xem đề đoạn "trộn 1 mol HCl" xem có nhầm lẫn gì không nhé.

 

 

\(n_{NaOH}=\dfrac{200.4\%}{40}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{200.7,3\%}{36,5}=0,4\left(mol\right)\\ NaOH+HCl\rightarrow NaCl+H_2O\\ m_{ddsau}=200+200=400\left(g\right)\\ Vì:\dfrac{0,2}{1}< \dfrac{0,4}{1}\Rightarrow HCldư\\ n_{HCl\left(p.ứ\right)}=n_{NaCl}=n_{NaOH}=0,2\left(mol\right)\\ n_{HCl\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\\ C\%_{ddHCl\left(dư\right)}=\dfrac{0,2.36,5}{400}.100=1.825\%\\ C\%_{ddNaCl}=\dfrac{0,2.58,5}{400}.100=2,925\%\)

NaCl+HCl→NaCl+H2O

nHCl=300.7,3%/36,5=0,6>nNaOH=200.4%/40=0,2→HCl dư

nHCl pư=nNaCl=nNaOH=0,2(mol)

nHCl dư=0,6−0,2=0,4(mol)

mdd sau pư=300+200=500(gam)

C%NaCl=0,2.58,5500.100%=2,34%

C%HCl=0,4.36,5500.100%=2,92%

Trộn 200g dd chi đó bn

a)

$NaOH + HCl \to NaCl + H_2O$

b)

Theo PTHH : 

$n_{HCl} = n_{NaOH} = \dfrac{80.20\%}{40}= 0,4(mol)$

$C\%_{HCl} = \dfrac{0,4.36,5}{200}.100\% = 7,3\%$

c)

$n_{NaCl} = 0,4(mol) \Rightarrow m_{NaOH} = 0,4.58,5 = 23,4(gam)$

a/ Bacl2 + H2So4 --> baso4 + 2hcl 0,2 <-- 0,2 --> 0,2 0,4 b/ nBacl2 = 0,86 mol nH2so4 = 0,2 molmuối thu được sau phản ứng là bacl2 dưmBacl2 = (0,86 -0,2) * 208= 137,28gc/ m dd sau pu= mBacl2 + mH2so4 - mBaso4 =200 + 200 - 0,2 * 233= 353.4g %Bacl2 dư = 137,28/353,4% =38,84% %HCl = (0,4 * 36,5)/353,4 % =4.131%

Làm sao ạ

\(n_{NaOH}=\dfrac{150.20\%}{40}=0,75\left(mol\right)\\ n_{HCl}=\dfrac{250.7,3\%}{36,5}=0,5\left(mol\right)\\ NaOH+HCl\rightarrow NaCl+H_2O\\ Vì:\dfrac{0,75}{1}>\dfrac{0,5}{1}\Rightarrow NaOHdư\\ \Rightarrow n_{NaOH\left(p.ứ\right)}=n_{NaCl}=n_{HCl}=0,5\left(mol\right)\\ n_{NaOH\left(dư\right)}=0,75-0,5=0,25\left(mol\right)\\ C\%_{ddNaCl}=\dfrac{58,5.0,5}{150+250}.100=7,3125\%\\ C\%_{ddNaOH\left(dư\right)}=\dfrac{0,25.40}{150+250}.100=2,5\%\)

a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)

b, \(n_{Mg}=\dfrac{1,2}{24}=0,05\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Mg}=0,1\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,1.36,5}{10\%}=36,5\left(g\right)\)

c, \(n_{H_2}=n_{MgCl_2}=n_{Mg}=0,05\left(mol\right)\)

Ta có: m _{dd sau pư} = 1,2 + 36,5 - 0,05.2 = 37,6 (g)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,05.95}{37,6}.100\%\approx12,63\%\)

\(a.n_{CO_2}=\dfrac{3,36}{22,4}=0,15mol\\ Na_2CO_3+2HCl\rightarrow2NaCl+CO_2\uparrow+H_2O\left(1\right)\\ NaOH+HCl\rightarrow NaCl+H_2O\\ n_{Na_2CO_3}=n_{CO_2}=0,15mol\\ \%m_{Na_2CO_3}=\dfrac{0,15.106}{23,9}\cdot100=66,5\%\\ \%m_{NaOH}=100-66,5=33,5\%\)

b. Sai đề, vì 

\(n_{HCl\left(thực,tế\right)}=\dfrac{200.3,65}{100}:36,5=0,2mol\\ n_{HCl\left(pư\right)}=0,15.2+\left(23,9-0,15.106\right):40=0,5mol\)

mà \(n_{CO_2}=0,15mol\Rightarrow n_{HCl}=0,3mol\left(pt1\right)\)(nên NaOH và Na₂CO₃ ko dư)

vậy cần ít nhất 0,5mol HCl để tính 

⇒cần thay đổi \(m_{ddHCl}\) hoặc \(C_{\%HCl}\) để tính được câu b

\(n_{Na_2O}=\dfrac{6.2}{62}=0.1\left(mol\right)\)

\(n_{HCl}=\dfrac{91.25\cdot10\%}{36.5}=0.25\left(mol\right)\)

\(Na_2O+2HCl\rightarrow2NaCl+H_2O\)

\(TC:\dfrac{0.1}{1}< \dfrac{0.25}{2}\Rightarrow HCldư\)

\(m_{NaCl}=0.1\cdot2\cdot58.5=11.7\left(g\right)\)

\(m_{dd}=6.2+91.25=97.45\left(g\right)\)

\(C\%_{NaCl}=\dfrac{11.7}{97.45}\cdot100\%=12\%\)

\(C\%_{HCl\left(dư\right)}=\dfrac{\left(0.25-0.2\right)\cdot36.5}{97.45}\cdot100\%=1.87\%\)

nNa2O=0,1(mol)

PTHH: Na2O + H2O -> 2 NaOH

-> nNaOH=0,2(mol)

nHCl=9,125(mol)->nHCl=0,25(mol)

PTHH: NaOH + HCl -> NaCl + H2O

Vì 0,25/1 > 0,2/1

=> NaOH hết, HCl dư, tính theo nNaOH

-> nNaCl=nHCl(p.ứ)=nNaOH=0,2(mol)

=>mNaCl=58,5.0,2= 11,7(g)

mHCl(dư)=0,05.36,5= 1,825(g)

mddsau=0,2.40+ 91,25= 99,25(g)

=>C%ddHCl(dư)=(1,825/99,25).100=1,839%

C%ddNaCl=(11,7/99,25).100=11,788%