* Giả sử n Ca(H2PO4)2=1 (mol) =n Ca trong Ca(H2PO4)2 = 4n H trong Ca(H2PO4)2 = 2n P trong Ca(H2PO4)2 = 8n O trong Ca(H2PO4)2

=> m Ca(H2PO4)2= 234 (g)

%m Ca trong Ca(H2PO4)2 = 40/234*100=17.09%

%m H trong Ca(H2PO4)2= 4/234*100=1.71%

%m P trong Ca(H2PO4)2= (2*31)/234*100=26,5%

%m O trong Ca(H2PO4)2= (8*16)/234*100=54.7%

*Cái (NH4)2HPO4 tương tụ nhé. Mà (NH4)2HPO là (NH4)2HPO4 chứ nhỉ?

Khối lượng  C a ( H ₂ P O ₄ ) 2  trong 15,55 g supephotphat đơn:

Khối lượng  P 2 O 5  trong mẫu supephotphat đơn trên :

1  C a ( H ₂ P O ₄ ) 2  - 1  P 2 O 5

% về khối lượng của  P 2 O 5 : 

a, \(\%m_{Al}=\dfrac{54.100\%}{102}=53\%\)

\(\%m_O=100\%-53\%=47\%\)

b, \(\%m_{Ca}=\dfrac{40.100\%}{232}=17,24\%\)

\(\%m_H=\dfrac{2.100\%}{232}=0,86\%\)

\(\%m_P=\dfrac{62.100\%}{232}=26,72\%\)

\(\Rightarrow\%m_O=100\%-17,24\%-0,86\%-26,72\%=55,18\%\)

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

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

   \(V_{H_2}=0,08\cdot22,4=1,792l\)

\(\%m_{Ca}=\dfrac{40}{164}.100=24,39\left(\%\right)\\ \%m_N=\dfrac{28}{164}.100=17,07\left(\%\right)\\ \%m_O=\dfrac{48.2}{164}.100=58,54\left(\%\right)\)

$M_{Ca(NO_3)_2} = 164$

$\%Ca = \dfrac{40}{164}.100\% = 24,39\%$
$\%N = \dfrac{14.2}{164}.100\% = 17,07\%$

$\%O = 100% -24,39\% -17,07\% = 58,54%%

Dễ ợt

\(1,\left\{{}\begin{matrix}p=e\\n+p+e=40\\2p-n=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2p+n=40\\2p-n=12\end{matrix}\right.\Leftrightarrow n=\dfrac{40-12}{2}=14\)

\(2,PTK_{Al_2O_3}=2\cdot27+16\cdot3=102\left(đvC\right)\\ PTK_{Al_2\left(SO_4\right)_3}=2\cdot27+\left(32+16\cdot4\right)\cdot3=342\left(đvC\right)\\ PTK_{Fe\left(NO_3\right)_3}=56+\left(14+16\cdot3\right)\cdot3=242\left(đvV\right)\\ PTK_{Na_3PO_4}=23\cdot3+31+16\cdot4=164\left(đvC\right)\\ PTK_{Ca\left(H_2PO_4\right)_2}=40+\left(2+31+16\cdot4\right)\cdot2=234\left(đvC\right)\\ PTK_{Ba_3PO_4}=137\cdot3+31+16\cdot4=506\left(đvC\right)\\ PTK_{ZnSO_4}=65+32+16\cdot4=161\left(đvC\right)\\ PTK_{AgCl}=108+35,5=143,5\left(đvC\right)\\ PTK_{NaBr}=23+80=103\left(đvC\right)\)

3/ ⁿhỗn hợp = 8,4.10²³ : 6.10²³ = 1,4 (mol)

ⁿO = 230,4 : 16 = 14,4 (mol)

Gọi ⁿCa₃(PO₄)₂ = x (mol) \(\rightarrow\) ⁿO = 8x (mol)

\(\rightarrow\) ⁿAl₂(SO₄)₃ = 1,4-x (mol) \(\rightarrow\) ⁿO = 12.(1,4-x) (mol)

\(\rightarrow\) 8x + 12.(1,4-x) = 14,4 \(\rightarrow\) x = 0,6 (mol)

ⁿCa₃(PO₄)₂= 0,6 (mol) \(\rightarrow Ca_3\left(PO_4\right)_2=\) 0,6.310 = 186 (g)

ⁿAl₂(SO₄)₃= 1,4-x = 0,8 (mol) \(\rightarrow^mAl_2\left(SO_4\right)_3\) = 0,8 . 342 = 273,6 (g)