\(a)n_{H_2SO_4}=\dfrac{58,8.20}{100.98}=0,12mol\\ n_{BaCl_2}=\dfrac{200.5,2}{100.208}=0,05mol\\ BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\\ \Rightarrow\dfrac{0,12}{1}>\dfrac{0,05}{2}\Rightarrow H_2SO_4.dư\\ BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)

0,05         0,05             0,05           0,1

\(m_{BaSO_4}=0,05.233=11,65g\\ b)m_{dd}=58,8+200-11,65=247,15g\\ C_{\%HCl}=\dfrac{0,1.36,5}{247,15}\cdot100=1,48\%\\ C_{\%H_2SO_4,dư}=\dfrac{\left(0,12-0,05\right).98}{247,15}\cdot100=2,78\%\)

mBaCl2=10.4(g)

nBaCl2=0.05(mol)

mH2SO4=11.76(g)

nH2SO4=0.12(mol)

BaCl2+H2SO4->BaSO4+2HCl

Theo pthh:nH2SO4=nBaCl2

theo bài ra,nH2SO4>nBaCl2

->H2SO4 dư

nH2SO4 dư=0.12-0.05=0.07(mol)

mH2SO4 dư=0.07*98=6.86(g)

nBaSO4=0.05(mol)

mBaSO4=11.65(g)

nHCl=0.05*2=0.1(mol)

mHCl=3.65(g)

mdd sau phản ứng:200+58.8-11.65=247.15(g)

C%(HCl)=3.65:247.15*100=1.48%

C%(H2SO4)=6.86:247.15*100=2.78%

PTHH: \(BaCl_2+H_2SO_4\rightarrow2HCl+BaSO_4\downarrow\)

a+b) Ta có: \(n_{BaCl_2}=\dfrac{400\cdot5,2\%}{208}=0,1\left(mol\right)=n_{H_2SO_4}=n_{BaSO_4}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ddH_2SO_4}=\dfrac{0,1\cdot98}{20\%}=49\left(g\right)\\m_{BaSO_4}=0,1\cdot233=23,3\left(g\right)\end{matrix}\right.\)

c) Theo PTHH: \(n_{HCl}=0,2\left(mol\right)\) \(\Rightarrow m_{HCl}=0,2\cdot36,5=7,3\left(g\right)\)

Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{ddBaCl_2}+m_{ddH_2SO_4}-m_{BaSO_4}=425,7\left(g\right)\)

\(\Rightarrow C\%_{HCl}=\dfrac{7,3}{425,7}\cdot100\%\approx1,71\%\)

Bạn xem lại giúp mình , coi đề có bị thiếu gì không nhé 

ai giúp mình với, mình đang ktra huhuhuh

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

\(a.BaCl_2+2AgNO_3\rightarrow Ba\left(NO_3\right)_3+2AgCl\\b. n_{AgNO_3}=\dfrac{300.20\%}{170}=\dfrac{6}{17}\left(mol\right)\\ n_{AgCl}=\dfrac{24}{143,5}=\dfrac{48}{287}\left(mol\right)\\ Tacó:n_{AgNO_3\left(pư\right)}=n_{AgCl}=\dfrac{48}{287}\left(mol\right)\\ \Rightarrow H=\dfrac{\dfrac{48}{287}}{\dfrac{6}{17}}.100=47,39\%\\ c.m_{ddsaupu}=300+200-24=476\left(g\right)\\ n_{Ba\left(NO_3\right)_2}=\dfrac{1}{2}n_{AgCl}=\dfrac{24}{287}\left(mol\right)\\ n_{AgNO_3\left(dư\right)}=\dfrac{6}{17}-\dfrac{48}{287}=\dfrac{906}{4879}\left(mol\right)\\ \Rightarrow C\%_{Ba\left(NO_3\right)_2}=4,59\%;C\%_{AgNO_3\left(dư\right)}=6,63\%\)

Tham khảo

a.BaCl2+2AgNO3→Ba(NO3)3+2AgClb.nAgNO3=300.20%170=617(mol)nAgCl=24143,5=48287(mol)Tacó:nAgNO3(pư)=nAgCl=48287(mol)⇒H=48287617.100=47,39%c.mddsaupu=300+200−24=476(g)nBa(NO3)2=12nAgCl=24287(mol)nAgNO3(dư)=617−48287=9064879(mol)⇒C%Ba(NO3)2=4,59%;C%AgNO3(dư)=6,63%

TK

https://hoc24.vn/cau-hoi/cho-300g-dung-dich-agno3-20-phan-ung-voi-200g-dung-dich-bacl2-thuduoc-ket-tua-loc-say-kho-ket-tua-can-nang-24ga-viet-pthhb-tinh-hieu-suat-cua-pha.3392944860763

Bài 19  :

\(a) n_{Al} = \dfrac{10,8}{27} = 0,4(mol)\\ 2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\\ n_{H_2} = \dfrac{3}{2}n_{Al} = 0,6(mol)\\ V_{H_2} = 0,6.22,4 = 13,44(lít)\\ b) \text{Chất tan : }Al_2(SO_4)_3\\ n_{Al_2(SO_4)_3} = \dfrac{1}{2}n_{Al} = 0,2(mol)\\ m_{Al_2(SO_4)_3} = 0,2.342 = 68,4(gam)\)

Bài 18 : 

\(a) n_{HCl} = \dfrac{250.7,3\%}{36,5 } = 0,5(mol)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,25(mol) \Rightarrow V_{H_2} = 0,25.22,4 = 5,6(lít)\\ b) \text{Chất tan : } ZnCl_2\\ n_{ZnCl_2} = n_{H_2} = 0,25(mol)\\ m_{ZnCl_2} = 0,25.136 = 34(gam)\)

m_ddH2SO4 = 100 . 1,137 = 113,7

n_H2SO4 = 113,7 . 20%/98 = 0,232 mol

n_BaCl2 = 400 . 5,29%/208 = 0,1 mol

                 H₂SO₄ + BaCl₂ —> BaSO₄ + 2HCI

Bđ:            0,232      0,1

Pứ:            0,1        0, 1         0,1       0,2

Sau pứ:      0,132         0

m_BaSO4 = 0,1.233 = 23,3 gam

Khối lượng dung dịch sau khi lọc bỏ kết tủa:

m_ddB = m_ddH2SO4 + m_ddBaCl2 - m_BaSO4 = 490,4

C%HCI = 0,2.36,5/490,4 = 1,49%

C%_H2SO4 dư = 0,132.98/490,4 = 2,64%