n_FeS2 = \(\dfrac{120000}{120}\) = 1000mol

2FeS₂ + 2H₂O + 7O₂ \(^{to}\rightarrow\) 2FeSO₄+ 2H₂SO₄

1000---------------------------------------->1000

=>m_{H2SO4(dung dịch)} = 1000 . 1,84 = 1840 g

=>n_H2SO4 = \(\dfrac{1840.98}{100.98}\) = 18,4 mol

=>m_H2SO4 = 18,4 . 98 = 1803,2 g = 1,8032 kg

mà H = 80%

=> m_{H2SO4(thực tế)} = 1,8032 . \(\dfrac{80}{100}\) = 1,44256 kg

- Giả sử có 100 gam dd H₂SO₄ 98%

\(m_{H_2SO_4}=\dfrac{100.98}{100}=98\left(g\right)\) => \(n_{H_2SO_4}=\dfrac{98}{98}=1\left(mol\right)\)

\(V_{dd.H_2SO_4.98\%}=\dfrac{100}{1,84}=\dfrac{1250}{23}\left(ml\right)=\dfrac{5}{92}\left(l\right)\)

\(C_{M\left(dd.H_2SO_4.98\%\right)}=\dfrac{1}{\dfrac{5}{92}}=18,4M\)

-

\(n_{H_2SO_4}=18,4.0,05=0,92\left(mol\right)\)

=> \(m_{H_2SO_4}=0,92.98=90,16\left(g\right)\)

=> \(m_{dd.H_2SO_4.10\%}=\dfrac{90,16.100}{10}=901,6\left(g\right)\)

=)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))

Cái đấy là Hóa mà đâu có phải là toán đâu Trong mục này là Toán lớp 2 nha

=))

thế mới hay

Phương trình hóa học phản ứng 

2FeS₂ + 2H₂O + 3O₂ ----> 2FeSO₄ + 2H₂SO₄ 

\(n_{FeS_2}=\frac{m}{M}=\frac{60000}{120}=500\left(mol\right)\)

\(n_{H_2SO_4\text{ lí thuyết }}=500\text{ mol}\)

=> \(m_{H_2SO_4}=n.M=500.98=49000\left(g\right)\)

=> \(V_{H_2SO_4}=\frac{m}{D}=\frac{49000}{1.84}=26630\left(ml\right)=26,63\left(l\right)\)

 => H_{điều chế} = \(\frac{25,8}{26,63}.100\%\approx97\%\)

P/S : Không chắc chắn lắm về bài làm !!!!

Bài 10:

- Giả sử có 100 gam dd H₂SO₄ 98%

\(m_{H_2SO_4}=\dfrac{100.98}{100}=98\left(g\right)\) => \(n_{H_2SO_4}=\dfrac{98}{98}=1\left(mol\right)\)

\(V_{dd.H_2SO_4.98\%}=\dfrac{100}{1,84}=\dfrac{1250}{23}\left(ml\right)=\dfrac{5}{92}\left(l\right)\)

\(C_{M\left(dd.H_2SO_4.98\%\right)}=\dfrac{1}{\dfrac{5}{92}}=18,4M\)

\(n_{H_2SO_4}=18,4.0,05=0,92\left(mol\right)\)

=> \(m_{H_2SO_4}=0,92.98=90,16\left(g\right)\)

=> \(m_{dd.H_2SO_4.10\%}=\dfrac{90,16.100}{10}=901,6\left(g\right)\)

Bài 11:

a) Do dd sau pư có 3 chát tan với nồng độ % bằng nhau

=> \(m_{Al_2\left(SO_4\right)_3}=m_{ZnSO_4}=m_{H_2SO_4\left(dư\right)}\)

Gọi số mol Al, Zn là a, b (mol)

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

             a----->1,5a------->0,5a----->1,5a

            Zn + H₂SO₄ --> ZnSO₄ + H₂

             b----->b--------->b----->b

=> \(\left\{{}\begin{matrix}m_{Al_2\left(SO_4\right)_3}=342.0,5a=171a\left(g\right)\\m_{ZnSO_4}=161b\left(g\right)\end{matrix}\right.\)

=> 171a = 161b 

=> \(\dfrac{a}{b}=\dfrac{161}{171}\) (1)

Có: \(\dfrac{m_{Al}}{m_{Zn}}=\dfrac{27.n_{Al}}{65.n_{Zn}}=\dfrac{27}{65}.\dfrac{161}{171}=\dfrac{483}{1235}\)

b) \(n_{H_2}=1,5a+b=\dfrac{11,2}{22,4}=0,5\left(mol\right)\) (2)

(1)(2) => \(\left\{{}\begin{matrix}a=\dfrac{161}{825}\left(mol\right)\\b=\dfrac{57}{275}\left(mol\right)\end{matrix}\right.\)

=> \(x=\dfrac{161}{825}.27+\dfrac{57}{275}.65=\dfrac{5154}{275}\left(g\right)\)

\(m_{H_2SO_4\left(dư\right)}=m_{Al_2\left(SO_4\right)_3}=342.0,5\dfrac{161}{825}=\dfrac{9177}{275}\left(g\right)\)

=> \(m_{H_2SO_4\left(bđ\right)}=98\left(1,5a+b\right)+\dfrac{9177}{275}=\dfrac{22652}{275}\left(g\right)\)

=> \(y=\dfrac{\dfrac{22652}{275}.100}{10}=\dfrac{45304}{55}\left(g\right)\)

Đáp án A

Số mol FeS₂ là:

Ta có: \(n_{FeS_2}=\dfrac{60}{120}=0,5\left(kmol\right)\)

Bảo toàn nguyên tố: \(n_{H_2SO_4\left(lý.thuyết\right)}=2n_{FeS_2}=1\left(kmol\right)\)

Mặt khác: \(n_{H_2SO_4\left(thực\right)}=\dfrac{25,8\cdot1,84\cdot98\%}{98}=0,47472\left(kmol\right)\)

\(\Rightarrow H\%=\dfrac{0,47472}{1}\cdot100\%=47,472\%\)

đề có thiếu ko bạn

mH2SO4=(3000.1,26).35%= 1323g

mddH2SO4=1323:98%=1350--> V=1350:1,84=733,69ml