Bài 1: Tính Khối Lượng Của Nguyên Tố Oxi có trong mỗi hợp chất sau:
1. 18 gam nước

\(n_{H_2O}=\dfrac{18}{18}=1\left(mol\right)\Rightarrow n_O=1.1=1\left(mol\right)\)

=> m_O = 1.16 = 16 (g)
2. 2,2 gam CO₂

\(n_{CO_2}=\dfrac{2,2}{44}=0,05\left(mol\right)\Rightarrow n_O=0,05.2=0,1\left(mol\right)\)

m_O = 0,1 .16 =1x6(g)

3. 8 gam CuSO₄

\(n_{CuSO_4}=\dfrac{8}{160}=0,05\left(mol\right)\Rightarrow n_O=0,05.4=0,2\left(mol\right)\)

=> m_O= 0,2.16= 3,2(g)

4. 2 gam Fe₂(SO₄)₃

\(n_{Fe_2\left(SO_4\right)_3}=\dfrac{2}{400}=0,005\left(mol\right)\Rightarrow n_O=0,005.12=0,06\left(mol\right)\)

=> m_O= 0,06.16 = 0,96(g)

a) 

Giả sử có 100 gam hỗn hợp

\(m_O=\dfrac{25.100}{100}=25\left(g\right)\)

=> \(n_O=\dfrac{25}{16}=1,5625\left(mol\right)\)

Mà n_O = 4.n_S

=> \(n_S=\dfrac{1,5625}{4}=\dfrac{25}{64}\left(mol\right)\)

\(\%m_S=\dfrac{\dfrac{25}{64}.32}{100}.100\%=12,5\%\)

b) Đề bài cho rồi mà bn :)

c) 

C1: %m_{kim loại} = \(100\%-12,5\%-25\%=62,5\%\)

=> m_{kim loại} = \(\dfrac{64.62,5}{100}=40\left(g\right)\)

C2:

\(m_S=\dfrac{64.12,5}{100}=8\left(g\right)\)

\(m_O=\dfrac{64.25}{100}=16\left(g\right)\)

=> m_{kim loại} = 64 - 8 - 16 = 40 (g)

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

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

   \(V_{H_2}=0,08\cdot22,4=1,792l\)

Đáp án D

m_X = m_C + m_H + m_O  + m_Na =  12 ( n N a 2 C O 3 + n C O 2 ) + 2 n H 2 O + 23.2. n N a 2 C O 3 + m O

Þ 12.(0,15 + 0,05) + 2.0,15 + 46.0,05 + m_O = 8,2

Þ m_O = 3,2 gam

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

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

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)

\(m_{SO_4^{2-}}=m_S=\dfrac{64\cdot20\%}{100\%}=12,8g\Rightarrow n_S=n_{SO_4^{2-}}=0,4mol\)

Khối lượng kim loại:

\(m=64-m_{SO_4^{2-}}=64-0,4\cdot96=25,6g\)