K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

 

\(n_{H_2SO_4}=\dfrac{200.4,9\%}{98}=0,1\left(mol\right)\\ H_2SO_4+BaCl_2\rightarrow BaSO_4+2HCl\\ a.n_{BaSO_4}=n_{BaCl_2}=n_{H_2SO_4}=0,1\left(mol\right)\\ n_{HCl}=2.0,1=0,2\left(mol\right)\\ C\%_{ddBaCl_2}=\dfrac{0,1.208}{300}.100\approx6,933\%\\ b.m_{ddsau}=200+300-0,1.233=476,7\left(g\right)\\ C\%_{ddHCl}=\dfrac{0,2.36,5}{476,7}.100\approx1,531\%\)

13 tháng 4 2023

a, \(BaCl_2+H_2SO_4\rightarrow2HCl+BaSO_{4\downarrow}\)

b, \(m_{BaCl_2}=200.2,08\%=4,16\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{4,16}{208}=0,02\left(mol\right)\)

\(m_{H_2SO_4}=300.9,8\%=29,4\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,02}{1}< \dfrac{0,3}{1}\), ta được H2SO4 dư.

Theo PT: \(\left\{{}\begin{matrix}n_{BaSO_4}=n_{H_2SO_4\left(pư\right)}=n_{BaCl_2}=0,02\left(mol\right)\\n_{HCl}=2n_{BaCl_2}=0,04\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,3-0,02=0,28\left(mol\right)\)

Ta có: m dd sau pư = m dd BaCl2 + m dd H2SO4 - mBaSO4 = 200 + 300 - 0,02.233 = 495,34 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{0,04.36,5}{495,34}.100\%\approx0,295\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,28.98}{495,34}.100\%\approx5,54\%\end{matrix}\right.\)

Bài 10:

PTHH: \(Na_2CO_3+BaCl_2\rightarrow2NaCl+BaCO_3\downarrow\)

a) Ta có: \(n_{Na_2CO_3}=\dfrac{200\cdot10,6\%}{106}=0,2\left(mol\right)=n_{BaCO_3}\)

\(\Rightarrow m_{BaCO_3}=0,2\cdot197=39,4\left(g\right)\)

b) Theo PTHH: \(n_{BaCl_2}=n_{BaCO_3}=0,2mol\)

\(\Rightarrow C\%_{BaCl_2}=\dfrac{0,2\cdot208}{120}\cdot100\%\approx34,67\%\)

c) Theo PTHH: \(n_{NaCl}=2n_{BaCl_2}=0,4mol\) \(\Rightarrow m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\)

Mặt khác: \(m_{dd}=m_{ddNa_2CO_3}+m_{ddBaCl_2}-m_{BaCO_3}=280,6\left(g\right)\)

\(\Rightarrow C\%_{NaCl}=\dfrac{23,4}{280,6}\cdot100\%\approx8,34\%\)

Bài 11 và bài 12 tương tự bài 10

 

18 tháng 11 2021

\(a,PTHH:H_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2HCl\\ \left\{{}\begin{matrix}m_{H_2SO_4}=\dfrac{300\cdot9,8\%}{100\%}=29,4\left(g\right)\\m_{BaCl_2}=\dfrac{200\cdot26\%}{100\%}=52\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\\n_{BaCl_2}=\dfrac{52}{208}=0,25\left(mol\right)\end{matrix}\right.\)

Vì \(\dfrac{n_{H_2SO_4}}{1}>\dfrac{n_{BaCl_2}}{1}\) nên H2SO4 dư

\(\Rightarrow n_{BaSO_4}=n_{BaCl_2}=0,25\left(mol\right)\\ \Rightarrow a=m_{BaSO_4}=0,25\cdot233=58,25\left(g\right)\\ b,n_{HCl}=n_{BaCl_2}=0,25\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,25\cdot36,5=9,125\left(g\right)\\ m_{dd_{HCl}}=300+200-58,25=441,75\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{9,125}{441,75}\cdot100\%\approx2,07\%\)

28 tháng 8 2018

a) Vì: mA < 400 (g) nên phải có khí thoát ra → muối có dạng MHSO4 và khí là: CO2

b)

c) Tác dụng được với: MgCO3, Ba(HSO3)2, Al2O3, Fe(OH)2, Fe, Fe(NO3)2

Pt: 2NaHSO4 + MgCO3 → Na2SO4 + MgSO4 + CO2↑ + H2O

2NaHSO4 + Ba(HSO3)2 → BaSO4 + Na2SO4 + SO2↑ + 2H2O

6NaHSO4 + Al2O3 → 3Na2SO4 + Al2(SO4)3 + 3H2O

2NaHSO4 + Fe(OH)2 → Na2SO4 + FeSO4 + 2H2O

2NaHSO4 + Fe → Na2SO4 + FeSO4 + H2

12NaHSO4 + 9Fe(NO3)2 → 5Fe(NO3)3 + 2Fe2(SO4)3 + 6Na2SO4 + 3NO↑ + 6H2O

9 tháng 11 2021

ai giúp mình với, mình đang ktra huhuhuh

 

9 tháng 11 2021

\(m_{H_2SO_4}=\dfrac{19,6\cdot20\%}{100\%}=3,92\left(g\right)\\ \Rightarrow n_{H_2SO_4}=\dfrac{3,92}{98}=0,04\left(mol\right)\\ PTHH:H_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2HCl\\ \Rightarrow n_{H_2SO_4}=n_{BaCl_2}=n_{BaSO_4}=0,04\left(mol\right)\\ \Rightarrow m_{CT_{BaCl_2}}=0,04\cdot208=8,32\left(g\right)\\ \Rightarrow m_{dd_{BaCl_2}}=\dfrac{8,32\cdot100\%}{12\%}\approx69,3\left(g\right)\\ m_{kết.tủa}=m_{BaSO_4}=0,04\cdot233=9,32\left(g\right)\)

18 tháng 11 2021

\(a,\left\{{}\begin{matrix}m_{BaCl_2}=\dfrac{100\cdot10,4\%}{100\%}=10,4\left(g\right)\\m_{H_2SO_4}=\dfrac{200\cdot9,8\%}{100\%}=19,6\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{BaCl_2}=\dfrac{10,4}{208}=0,05\left(mol\right)\\n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\end{matrix}\right.\)

\(PTHH:BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)

Vì \(\dfrac{n_{BaCl_2}}{1}< \dfrac{n_{H_2SO_4}}{2}\) nên \(H_2SO_4\) dư

\(\Rightarrow n_{BaSO_4}=n_{BaCl_2}=0,05\left(mol\right)\\ \Rightarrow m_{BaSO_4}=0,05\cdot233=11,65\left(g\right)\)

\(b,n_{HCl}=n_{BaSO_4}=0,05\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,05\cdot36,5=1,825\left(g\right)\\ \Rightarrow m_{dd_{HCl}}=100+200-11,65=288,35\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{1,825}{288,35}\cdot100\%\approx0,63\%\)

30 tháng 11 2021

\(\begin{cases} m_{H_2SO_4}=\dfrac{100.19,6\%}{100\%}=19,6(g)\\ m_{BaCl_2}=\dfrac{300.20,8\%}{100\%}=62,4(g) \end{cases} \Rightarrow \begin{cases} n_{H_2SO_4}=\dfrac{19,6}{98}=0,2(mol)\\ n_{BaCl_2}=\dfrac{62,4}{208}=0,3(mol) \end{cases}\\ a,PTHH:BaCl_2+H_2SO_4\to BaSO_4\downarrow +2HCl\)

Vì \(\dfrac{n_{H_2SO_4}}{1}<\dfrac{n_{BaCl_2}}{1}\) nên \(BaCl_2\) dư

\(\Rightarrow n_{BaSO_4}=0,2(mol)\\ \Rightarrow m_{BaSO_4}=0,2.233=46,6(g)\)

\(b,n_{HCl}=n_{BaSO_4}=0,2(mol)\\ \Rightarrow m_{CT_{HCl}}=0,2.36,5=7,3(g)\\ m_{dd_{HCl}}=100+300-46,6=353,4(g)\\ \Rightarrow C\%_{HCl}=\dfrac{7,3}{353,4}.100\%\approx 2,07\%\)

\(n_{BaCl_2}=\dfrac{208.10\%}{208}=0,1\left(mol\right)\\ a,BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ 0,1............0,1..............0,1.............0,2\left(mol\right)\\ b,m_{ddH_2SO_4}=\dfrac{0,1.98.100}{8}=122,5\left(g\right)\\ c,m_{kt}=m_{BaSO_4}=0,1.233=23,3\left(g\right)\\ d,m_{ddsau}=208+122,5-23,3=307,2\left(g\right)\\ C\%_{ddHCl}=\dfrac{0,2.36,5}{307,2}.100\approx2,376\%\)

30 tháng 9 2021

thanks you cậu

 

 

26 tháng 2 2023

\(n_{Na_2SO_4}=\dfrac{100.28,4\%}{142}=0,2\left(mol\right)\)

PTHH: \(Na_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2NaCl\) 

               0,2------>0,2--------->0,2

`=>` \(\left\{{}\begin{matrix}C\%_{BaCl_2}=\dfrac{0,2.208}{200}.100\%=20,8\%\\m=m_{BaSO_4}=0,2.233=46,6\left(g\right)\end{matrix}\right.\)