Có \(M_{A_3XO_4}=164đvC\)

\(\rightarrow3M_A+M_X+4M_O=164\)

\(\rightarrow3M_A+M_X=164-64=100\)

Xét \(M_A=23\) 

\(\rightarrow3.23+M_X=100\)

\(\rightarrow M_X=100-69=31đvC\)

Do vậy X là Photpho

Vậy CTHH là \(Na_3PO_4\)

1. \(m_{KOH}=500.3\%=15g\)

Vì KOH không bị điện phân nên nước tham gia điện phân

Ở Anot thu được khí Oxi

\(n_{O_2}=\frac{67,2}{22,4}=3mol\)

\(2H_2O\rightarrow2H_2+O_2\)

\(\rightarrow n_{H_2}=2n_{O_2}=6mol\)

\(\rightarrow V_{\text{khí}}=22,4.\left(6+3\right)=201,6l\)

2. \(m_{ddsaupu}=500-6.2-3.32=392g\)

\(\rightarrow C\%_{KOH}=\frac{15.100}{392}=3,82\%\)

\(2H_2O\rightarrow^{đpdd}2H_2+O_2\)

Ở Catot: \(2H^++2e\rightarrow H_2\)

Ở Anot: \(2O^{2-}\rightarrow O_2+4e\)

a) \(n_{Al}=\frac{5,4}{27}=0,2\left(mol\right)\)

PTHH : \(2Al+3H_2SO_4-->Al_2\left(SO_4\right)_3+3H_2\)    

Theo pthh : \(n_{H_2}=\frac{3}{2}n_{Al}=0,3\left(mol\right)\)

=> \(V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\)

b) Theo pthh :

\(n_{Al_2\left(SO_4\right)_3}=\frac{1}{2}n_{Al}=0,1\left(mol\right)\)

 \(n_{H_2SO_4\left(pứ\right)}=\frac{3}{2}n_{Al}=0,3\left(mol\right)\) => \(V_{H_2SO_4}=\frac{\left(0,3+0,03\right)}{1}=0,33\left(l\right)\)

=> \(n_{H_2SO_4\left(dư\right)}=0,3\cdot0,1=0,03\left(mol\right)\)

=> \(\hept{\begin{cases}C_{M\left(Al_2\left(SO_4\right)_3\right)}=\frac{0,1}{0,33}\approx0,3\left(M\right)\\C_{M\left(H_2SO_4\right)}=\frac{0,03}{0,33}\approx0,09\left(M\right)\end{cases}}\)

a/ \(S+O_2\rightarrow\left(t^o\right)SO_2\)

b/  \(n_S=4,8:32=0,15\left(mol\right)\)

Theo PTHH \(n_{SO_2}=n_S=0,15\left(mol\right)\)

\(\rightarrow V_{SO_2}=0,15.22,4=3,36\left(l\right)\)

c/Theo PTHH \(n_{O_2}=n_S=0,15\left(mol\right)\)

\(\rightarrow m_{O_2}=0,15.32=4,8\left(g\right)\)

vào link  https://lazi.vn/uploads/edu/answer/1605437929_lazi_5fb109e9b7a9a.jpg

ko gửi ảnh đc

gfvfvfvfvfvfvfv555