RCl2+2KOH\(\rightarrow\)R(OH)2+2KCl
RCl3+3KOH\(\rightarrow\)R(OH)3+3KCl
Ta có nRCl2=nR(OH)2
mR(OH)2=19,8g \(\rightarrow\) nR(OH)2= \(\frac{19,8}{R+34}\) mol
mRCl= 0,5R \(\rightarrow\) nRCl=\(\frac{0,5R}{\text{R+35,5}}\)
\(\rightarrow\)Ta có Pt \(\frac{\text{0,5R}}{\text{R+35,5}}\)=\(\frac{19,8}{R+34}\)
\(\Leftrightarrow\)0,5R(R+34)=19,8(R+35,5)
\(\Leftrightarrow\) 0,5R2-2,8R-702,9=0
\(\Leftrightarrow\) R=40 \(\rightarrow\) R là Ca
RCl2+2KOH→→R(OH)2+2KCl
RCl3+3KOH→→R(OH)3+3KCl
Ta có nRCl2=nR(OH)2
mR(OH)2=19,8g →→ nR(OH)2= 19,8R+3419,8R+34 mol
mRCl= 0,5R →→ nRCl=0,5RR+35,50,5RR+35,5
→→Ta có Pt 0,5RR+35,50,5RR+35,5=19,8R+3419,8R+34
⇔⇔0,5R(R+34)=19,8(R+35,5)
⇔⇔ 0,5R2-2,8R-702,9=0
⇔⇔ R=40 →→ R là Ca
