1. \(R+H_2O\rightarrow ROH+\dfrac{1}{2}H_2\)
\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(n_{H_2}=0,1mol\rightarrow n_R=0,1:\dfrac{1}{2}=0,2\left(mol\right)\)
\(M_R=\dfrac{m}{n}=\dfrac{7,8}{0,2}=39\left(g/mol\right)\)
Vậy R là Kali (K)
2. K + \(H_2O\rightarrow KOH+\dfrac{1}{2}H_2\)
-----------------------0,2--------0,1(mol)
\(m_{KOH}=n.M=0,2.56=11,2\left(g\right)\)
\(C\%_{KOH}=\dfrac{m_{KOH}.100}{m_{dd}}=\dfrac{11,2.100}{300}\simeq3,73\left(\%\right)\)
1.PTHH: \(2R+2H_2O\rightarrow2ROH+H_2\uparrow\)
...............0,2....................0,2...............0,1...(mol)
\(nH_2=\dfrac{2.24}{22.4}=0,1\left(mol\right)\)
\(M_R=\dfrac{7,8}{0,2}=39\)
Suy ra R là K
2. \(mROH=0,2\times56=11,2\left(g\right)\)
\(C\%=\dfrac{11,2}{300}\times100=3,73\%\)
a,ta có pthh là \(2R+2H_2O\rightarrow2ROH+H_2\)
nH2=2,24/22,4=0,1 Mol=>nR=0,2 Mol
=> M R=7,8/0,2=39g/Mol
R là K
b, C%=7,8/300.100%=2,6%
