Làm gì có Fe(SO3)4

làm gì có chất \(Fe_2\left(SO_3\right)_4\)

Câu a.

\(M_{Ca\left(NO_3\right)_2}=164\)g/mol

\(m_{Ca\left(NO_3\right)_2}=0,3\cdot164=49,2g\)

\(\%Ca=\dfrac{40}{164}\cdot100\%=24,39\%\)

\(m_{Ca}=\%Ca\cdot49,2=12g\)

\(\%N=\dfrac{14\cdot2}{164}\cdot100\%=17,07\%\)

\(m_N=\%N\cdot49,2=8,4g\)

\(m_O=49,2-12-8,4=28,8g\)

Các câu sau em làm tương tự nhé!

a)\(n_{Ca\left(NO_3\right)_2}=0,3mol\)

   \(n_{Ca}=n_{Ca\left(NO_3\right)_2}=0,3mol\)

   \(m_{Ca}=0,3\cdot40=12g\)

   \(n_N=2n_{Ca\left(NO_3\right)_2}=2\cdot0,3=0,6mol\)

   \(m_N=0,6\cdot14=8,4g\)

   \(n_O=6n_{Ca\left(NO_3\right)_2}=6\cdot0,3=1,8mol\)

   \(m_O=1,8\cdot16=28,8g\)

b)\(n_O=\dfrac{9,6}{16}=0,6mol\)

   Mà \(n_O=12n_{Fe_2\left(SO_4\right)_3}\Rightarrow n_{Fe_2\left(SO_4\right)_3}=\dfrac{0,6}{12}=0,05mol\)

   \(\Rightarrow m=20g\)

c)\(n_{CuSO_4}=\dfrac{3,2}{160}=0,02mol\)

   \(n_O=4n_{CuSO_4}=0,08mol=n_{H_2}\)

   \(V_{H_2}=0,08\cdot22,4=1,792l\)

Đặt tỉ lệ số phân tử Fe2(SO4)3 và MgSO4 là x,y 

=> Số nguyên tử O: 12x + 4y 

Tổng số nguyên tử: 17x +6y 

Ta có:

 \(\dfrac{12x+4y}{17x+6y}=\dfrac{32}{47}\Leftrightarrow47.\left(12x+4y\right)=32.\left(17x+6y\right)\\ \Leftrightarrow564x-544x=192y-188y\\ \Leftrightarrow20x=4y\\ \Leftrightarrow\dfrac{x}{y}=\dfrac{1}{5}\)

=> Phần trăm khối lượng mỗi chất trong hh:

\(\%mFe2\left(SO4\right)3=\dfrac{400.1}{400.1+120.5}.100=40\%\\ \%mMgSO4=100\%-40\%=60\%\)

nS = 8/32 = 0,25 (mol)

nFe2(SO4)3 = 0,25/3 = 1/12 (mol)

=> nO2 = 1/12 (mol)

PTHH: 2KMnO4 -> (t°) K2MnO4 + MnO2 + O2

nKMnO4 = 1/12 . 2 = 1/6 (mol)

mKMnO4 = 1/6 . 158 = 79/3 (g)

\(a,n_{\left(NH_4\right)_3PO_4}=0,6\left(mol\right)\\ \Rightarrow n_N=0,6.3=1,8\left(mol\right)\Rightarrow m_N=1,8.14=25,2\left(g\right)\\ n_H=4.3.0,6=7,2\left(mol\right)\Rightarrow m_H=7,2.1=7,2\left(g\right)\\ n_P=n_{hc}=0,6\left(mol\right)\Rightarrow m_P=0,6.31=18,6\left(g\right)\\ n_O=4.0,6=2,4\left(mol\right)\Rightarrow m_O=2,4.16=38,4\left(g\right)\)

\(b,n_S=\dfrac{6,4}{32}=0,2\left(mol\right)\Rightarrow n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}.0,2=\dfrac{1}{15}\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=342.\dfrac{1}{15}=22,8\left(g\right)\\ c,n_{Al_2\left(SO_4\right)_3}=\dfrac{20,52}{342}=0,06\left(mol\right)\\ n_O=4.3.0,06=0,72\left(mol\right)\\ \Rightarrow n_{CO_2}=\dfrac{0,72}{2}=0,36\left(mol\right)\Rightarrow V_{CO_2\left(đktc\right)}=0,36.22,4=8,064\left(l\right)\)

Câu1) n_CO2 =m/M=11/44=0,25(mol)

         nH₂= 9.10²³/6.10²³=1,5(mol)

     V_H2 =n.22,4=1,5.22,4=33,6(l)