\(2Fe_2S+8SO_2\rightarrow8SO_3+8H_2SO_4\)

Theo tỉ lệ khối lượng: 240 tấn tạo 784 tấn

=> 12 tấn tạo \(\frac{12.784}{240}=39,2\) tấn

Thực tế tạo ra 30.49%= 14,7 tấn

Bạn ơi đề là dấu' \(\rightarrow\) ' chứ đâu phải dấu +

\(n_{FeS_2}=\frac{12000000}{120}=100000\left(mol\right)\)

\(sdpu:FeS_2\rightarrow SO_2\rightarrow SO_3\rightarrow H_2SO_4\)

(mol)______ \(1\rightarrow2\rightarrow2\rightarrow2\)______

Ta thấy: 1 mol FeS₂ sinh ra 2 mol H₂SO₄

Do đó: 100000 mol FeS₂ sẽ sinh ra 200000 mol H₂SO₄

\(\Rightarrow m_{H_2SO_4}=200000.98=19600000\left(g\right)=19,6\left(tan\right)\)

\(H=\frac{30.49\%}{19,6}.100\%=75\left(\%\right)\)

\(n_{H_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\\ m_{H_2}=1,5.2=3\left(g\right)\)

PTHH : 2Al + H₂SO₄ -> Al₂SO₄ + H₂

Theo ĐLBTKL

\(m_{Al}+m_{H_2SO_4}=m_{Al_2SO_4}+m_{H_2}\\ \Rightarrow m_{H_2SO_4}=\left(171+3\right)-2,7=171,3\left(g\right)\)

pthh: 2Al+3H\(_2\)SO\(_4\)→Al\(_2\)(SO4)\(_3\)+3H\(_2\)↑

nH\(_2=33,6:22,4=1,5\left(mol\right)\)

\(mH_2=1,5.2=3\left(g\right)\)

\(nAl_2\left(SO_4\right)=171:150=1,14\left(mol\right)\)

\(mAl_2\left(SO_4\right)_3=1,14.342=389,88\left(g\right)\)

BTKL : mAl + mH\(_2\)SO\(_4\) = m Al\(_2\)(SO4)\(_3\)  + m H\(_2\)

           2,7    +  mH\(_2\)SO\(_4\) =  389,88       +   3

=> \(mH_2SO_4=\left(389,88+3\right)-2,7=390,18\left(g\right)\)  

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(3Zn+2H_3PO_4\rightarrow Zn_3\left(PO_4\right)_2+3H_2\)

a)

\(BaCl_2+K_2SO_4\rightarrow BaSO_4+2KCl\\ BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)

b)

\(Ba\left(NO_3\right)_2+Na_2SO_4\rightarrow BaSO_4+2NaNO_3\\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

a) \(Pt:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b) \(n_{Fe}=\dfrac{0,56}{56}=0,01mol\)

Theo pt: \(n_{FeSO_4}=n_{Fe}=0,01mol\)

\(\Rightarrow m_{FeSO_4}=0,01.152=1,52g\)

Theo pt: \(n_{H_2}=n_{Fe}=0,01mol\)

\(\Rightarrow V_{H_2}=0,01.22,4=0,224lít\)

c) \(Theopt:nH_2SO_4=n_{Fe}=0,01mol\)

\(\Rightarrow m_{H_2SO_4}=0,01.98=0,98g\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,98.100}{19,6}=5g\)