%m_{Fe ( trong A )} = $\frac{112}{160}.60=42\%$

=> m_{Fe ( trong A )} = $\frac{42}{100}.1=0,42\left(\text{tấn}\right)=420\left(kg\right)$

Vậy trong 1 tấn quặng A có chứa 420 kg Fe

%m_{Fe ( trong B )} = $\frac{168}{232}.69,6=50,4\%$

=> m_{Fe ( trong B )} = $\frac{50,4}{100}.1=0,504\left(\text{tấn}\right)=504\left(kg\right)$

Vậy trong 1 tấn quặng B có chứa 504 kg Fe

%m_Fe2O3 = $\frac{3}{10}.100=30\%$

%m_{Fe3O4 = $\frac{7}{10}.100=70\%$}

=> m_{Fe( quặng A trong C )} = $\frac{30.420}{100}=126\left(kg\right)$

m_{Fe ( quặng B trong C )} =$\frac{70.504}{100}=352,8\left(kg\right)$

=> m_{Fe ( trong C )} = 126 + 352,8 = 478,8 (kg)

\(m_{Fe}=1000.95\%=950(kg)\\ \Rightarrow n_{Fe}=\dfrac{950}{56}=\dfrac{475}{28}(kmol)\\ PTHH:Fe_2O_3+3CO\xrightarrow{t^o}2Fe+3CO_2\\ \Rightarrow n_{Fe_2O_3}=2n_{Fe}=\dfrac{475}{14}(kmol)\\ \Rightarrow m_{Fe_2O_3}=\dfrac{475}{14}.160=5428,57(kg)\\ \Rightarrow m_{quặng}=\dfrac{5428,57}{90\%}\approx6030(kg)=6,03(tấn)\\ n_{CO_2}=\dfrac{3}{2}n_{Fe}=\dfrac{1425}{56}(kmol)\\ \Rightarrow m_{CO_2}=\dfrac{1425}{56}.44\approx1100(kg)=1,1(tấn)\)

Đặt m_A = a (tấn); m_B = b (tấn)

Giả sử a + b = 1 (tấn) (1)

\(m_{Fe_2O_3\left(A\right)}=a.60\%=0,6a\left(tấn\right)=6.10^5a\left(g\right)\)

=> \(n_{Fe_2O_3\left(A\right)}=\dfrac{6.10^5a}{160}=3750a\left(mol\right)\Rightarrow n_{Fe\left(A\right)}=7500a\left(mol\right)\)

\(m_{Fe_3O_4\left(B\right)}=b.69,6\%=0,696b\left(tấn\right)=696.10^3b\left(g\right)\)

=> \(n_{Fe_3O_4\left(B\right)}=\dfrac{696.10^3b}{232}=3000b\left(mol\right)\Rightarrow n_{Fe\left(B\right)}=9000b\left(mol\right)\)

\(n_{Fe\left(tổng\right)}=\dfrac{0,48.10^6}{56}=\dfrac{60000}{7}\left(mol\right)\)

=> \(7500a+9000b=\dfrac{60000}{7}\) (2)

(1)(2) => \(a=\dfrac{2}{7}\left(tấn\right);b=\dfrac{5}{7}\left(tấn\right)\)

=> \(\dfrac{a}{b}=\dfrac{2}{5}\)

Đặt m_A = a (tấn); m_B = b (tấn)

Giả sử a + b = 1 (tấn) (1)

\(m_{Fe_2O_3\left(A\right)}=a.60\%=0,6a\left(tấn\right)=6.10^5a\left(g\right)\)

=> \(n_{Fe_2O_3\left(A\right)}=\dfrac{6.10^5a}{160}=3750a\left(mol\right)\Rightarrow n_{Fe\left(A\right)}=7500a\left(mol\right)\)

\(m_{Fe_3O_4\left(B\right)}=b.69,6\%=0,696b\left(tấn\right)=696.10^3b\left(g\right)\)

=> \(n_{Fe_3O_4\left(B\right)}=\dfrac{696.10^3b}{232}=3000b\left(mol\right)\Rightarrow n_{Fe\left(B\right)}=9000b\left(mol\right)\)

\(n_{Fe\left(tổng\right)}=\dfrac{0,48.10^6}{56}=\dfrac{60000}{7}\left(mol\right)\)

=> \(7500a+9000b=\dfrac{60000}{7}\) (2)

(1)(2) => \(a=\dfrac{2}{7}\left(tấn\right);b=\dfrac{5}{7}\left(tấn\right)\)

=> \(\dfrac{a}{b}=\dfrac{2}{5}\)

1, \(n_{CuSO_4}=\dfrac{48}{160}=0,3 (mol)\)

\(n_{Cu}=n{CuSO_4}=0,3 mol\)

\(=> m_{Cu}=0,3 \) x 64 = 19, 2 (gam)

2, CHỊU

\(n=\dfrac{m}{M}=\dfrac{48}{160}=0,3\left(mol\right)\)

\(=>m=0,3.64=19,2\left(g\right)\)

Đổi 1 tấn = 1000 kg

nFe2O3 = 1000 . 90% : 160 = 5,625 (kmol)

nFe = 5,625 . 3 = 16,875 (kmol)

mFe = 16,875 . 945 (kg)

a. n_H2 = n_Fe = 0,1mol

Bảo toàn nguyên tố Fe => n_Fe2O3 = 0,05mol

=> m_Fe2O3 = 8g

=> %Fe₂O₃ = (8:10) . 100% = 80%

→ m F e 2 O 3   =   ( 1 , 68 . 160 )   :   112   =   2 , 4   t ấ n .