PTHH: \(Fe_2O_3+3CO\xrightarrow[]{t^o}2Fe+3CO_2\uparrow\)

Đổi 1 tấn = 1000 kg

Ta có: \(n_{Fe}=\dfrac{1000\cdot98\%}{56}=17,5\left(kmol\right)\) 

\(\Rightarrow n_{CO}=26,25\left(kmol\right)\) \(\Rightarrow V_{CO}=26,25\cdot22,4=588\left(m^3\right)\)

\(m_{Fe}=70\%.60\%.10=4,2\left(tấn\right)\)

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

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

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

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

=> \(\dfrac{a}{b}=\dfrac{2}{5}\)

Giúp e vs ạ. Cần gấp 

Ủa SAO LẠI CÓ CACBON Ở ĐÂY?

%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)

a) 2Fe(OH)₃ \(\underrightarrow{t^0}\) Fe₂O₃ + 3H₂O

b) n_Fe2O3 : n_H2O = 1 : 3 

c) n_Fe2O3 = 16/160 = 0.1 (mol) 

=> n_Fe(OH)3 = 0.2 * 107 = 21.4 g 

bt thì cc j