ai giúp mình với, mình đang ktra huhuhuh

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

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

n_Fe = 11.2/56 = 0.2 (mol) 

Fe + H2SO4 => FeSO4 + H2 

0.2____0.2_______0.2___0.2 

mH2SO4 = 0.2*98 = 19.6 (g) 

mdd H2SO4 = 19.6*100/10 = 196 (g) 

m dd sau phản ứng = 11.2 + 196 - 0.4 = 206.8 (g) 

mFeSO4 = 0.2*152 = 30.4 (g) 

C% FeSO4 = 30.4/206.8 * 100% = 14.7% 

Vdd H2SO4 = mdd H2SO4 / D = 196 / 1.14 = 171.9 (ml) 

CM FeSO4 = 0.2 / 0.1719 = 1.16 M 

n_Fe = 11.2/56 = 0.2 (mol) 

Fe + H2SO4 => FeSO4 + H2 

0.2____0.2_______0.2___0.2 

mH2SO4 = 0.2*98 = 19.6 (g) 

mdd H2SO4 = 19.6*100/10 = 196 (g) 

m dd sau phản ứng = 11.2 + 196 - 0.4 = 206.8 (g) 

mFeSO4 = 0.2*152 = 30.4 (g) 

C% FeSO4 = 30.4/206.8 * 100% = 14.7% 

Vdd H2SO4 = mdd H2SO4 / D = 196 / 1.14 = 171.9 (ml) 

CM FeSO4 = 0.2 / 0.1719 = 1.16 M 

a) \(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)

b) \(n_{BaCl_2}=\dfrac{52}{208}=0,25\left(mol\right)\)

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

            0,25----->0,25------->0,25---->0,5

=> \(m_{H_2SO_4}=0,25.98=24,5\left(g\right)\)

=> \(m_{ddH_2SO_4}=\dfrac{24,5.100}{19,6}=125\left(g\right)\)

c) \(m_{BaSO_4}=0,25.233=58,25\left(g\right)\)

d) 

\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)

PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)

Xét tỉ lệ \(\dfrac{0,5}{1}>\dfrac{0,2}{1}\) => NaOH hết, HCl dư

=> Quỳ tím chuyển màu đỏ

nAl = 5,4 / 27 = 0,2(mol)

2Al + 3H2SO4--- > Al2(SO4)3 + 3H2

0,2      0,3                0,1                0,3    (mol)

VH2SO4  = n/ CM = 0,3 / 2 = 0,15(l)

=> V1 = 150 ml

mAl2(SO4)3 = 0,1 . 342 = 34,2 (g)

Al2(SO4)3 + 3BaCl2 -- > 3BaSO4 + 2AlCl3

 0,1                                       0,1

=> mBaSO4 = 0,1 . 233 = 23,3 (g)

PTHH: \(NaCl+AgNO_3\rightarrow NaNO_3+AgCl\downarrow\)

Ta có: \(n_{NaCl}=0,2\cdot0,5=0,1\left(mol\right)=n_{AgNO_3}=n_{AgCl}\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{AgNO_3}}=\dfrac{0,1}{0,4}=0,25\left(M\right)\\m_{AgCl}=0,1\cdot143,5=14,35\left(g\right)\end{matrix}\right.\)

Cho 28 gam chất gì em nhỉ? Sắt hả?