Question

[Nhân Tư]

Hóa :v

\(mHCl_{\left(dd\right)}=\frac{a\cdot C}{100}\Leftrightarrow nHCl_{\left(dd\right)}=\frac{a\cdot C}{100\cdot36,5}=\frac{a\cdot C}{3650}\) 
\(\Leftrightarrow mH2O_{\left(dd\right)}=mdd-mHCl_{\left(dd\right)}=a-\frac{a\cdot C}{100}=\frac{\left(100-C\right)a}{100}\)

\(nH2=\frac{0,05a}{2}=0,025a\)
PT: 2Na + 2HCl ---> 2NaCl +H2 (1)
PT: Fe +2HCl ---->FeCl2+H2 (2)
PT: Na dư +H2O ----> NaPH +2 (3)
Theo (2) (3) => nH2=nHCl/2\(=\frac{a\cdot C}{3650\cdot2}=\frac{a\cdot C}{7300}\) (*)
Theo (3)(*) => \(\frac{\left(100-C\right)a}{1800}+\frac{aC}{7300}=nH2=0,025a\)
\(\Leftrightarrow\frac{\left(100-C\right)}{1800}+\frac{C}{7300}=0,025\)
=> C=73 (chẵn số)
=> C%= 73%

Answer 1

[Nhân Tư]

Bổ sung :v

Theo (3) nH2O=nH2=\(\frac{mH2O}{18}=\frac{\left(100-C\right)a}{18\cdot100}\)