Bao nhiêu g dd \(H_2SO_4\) 49% vậy?

a) Sửa đề: dd H₂SO₄ 9,8%

Ta có: \(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\) \(\Rightarrow m_{H_2}=0,35\cdot2=0,7\left(g\right)\)

Bảo toàn nguyên tố: \(n_{H_2SO_4}=n_{H_2}=0,35\left(mol\right)\) \(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,35\cdot98}{9,8\%}=350\left(g\right)\)

\(\Rightarrow m_{dd}=m_{KL}+m_{H_2SO_4}-m_{H_2}=361,6\left(g\right)\)

b) Tương tự câu a

Gọi A là kim loại có mặt trong oxit cần tìm

\(PTHH:AO+H_2SO_4\rightarrow ASO_4+H_2O\\ n_{H_2SO_4}=\dfrac{100.9,8\%}{98}=0,1\left(mol\right)\\ n_{AO}=n_{H_2SO_4}=0,1\left(mol\right)\\ M_{AO}=\dfrac{8}{0,1}=80\left(\dfrac{g}{mol}\right)=M_A+16\\ \Rightarrow M_A=64\left(\dfrac{g}{mol}\right)\\ \Rightarrow Oxit.AO:CuO\)

Gọi cthh của oxit kim loại hóa trị II là RO.

RO + H2SO4 --> RSO4 + H2O (1)

mH2SO4 = 9,8%.100 = 9,8 (g)

-> nH2SO4 = 9,8/98 = 0,1 (mol)

nRO = 8/R+16 (mol)

Từ (1) -> nRO = nH2SO4 = 0,1mol

-> 8/R+16 = 0,1 -> R = 64 -> R là Cu

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

nH2= 3,36/22,4= 0,15(mol)

=> nAl= 2/3 . nH2= 2/3 . 0,15 = 0,1(mol)

=> m= mAl= 0,1. 27= 2,7(g)

b) nH2SO4= nH2= 0,15(mol)

=> mH2SO4= 0,15. 98= 14,7(g)

Mà không có C% sao tính mdd đây?

\(n_{H_2SO_4}=\dfrac{300.9,8\%}{98}=0,3\left(mol\right)\\ BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ n_{HCl}=0,3.2=0,6\left(mol\right)\\ n_{BaSO_4}=n_{H_2SO_4}=0,3\left(mol\right)\\ m_{ddsau}=m_{ddBaCl_2}+m_{ddH_2SO_4}-m_{BaSO_4}=200+300-233.0,3=430,1\left(g\right)\\ C\%_{ddHCl}=\dfrac{0,6.36,5}{430,1}.100\approx5,092\%\)

\(n_{Fe_2O_3}=\dfrac{48}{160}=0,3\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂SO₄ → Fe₂(SO₄)₃ + 3H₂O

Mol:      0,3            0,9               0,3

\(m_{H_2SO_4}=0,9.98=88,2\left(g\right)\)

\(m_{ddH_2SO_4}=\dfrac{88,2}{10}.100=882\left(g\right)\)

m_{dd sau pứ} = 48+882 = 930 (g)

\(C\%_{ddFe_2\left(SO_4\right)_3}=\dfrac{0,3.400.100\%}{930}=30\%\)

\(n_{Fe_2O_3}=\dfrac{48}{160}=0,3\left(mol\right)\\ Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ 0,3........0,9.............0,3............0,9\left(mol\right)\\ m=m_{ddH_2SO_4}=\dfrac{0,9.98.100}{10}=882\left(g\right)\\ m_{ddsau}=48+882=930\left(g\right)\\ C\%_{ddsau}=C\%_{ddFe_2\left(SO_4\right)_3}=\dfrac{0,3.400}{930}.100\approx12,903\%\)