a, 2Na + 2H2O -> 2NaOH + H2
b, \(n_{H_2}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo pthh trên ta có: \(n_{Na}=2.0,2=0,4\left(mol\right)\)
=>\(m_{Na}=n.M=0,4.23=9,2\left(g\right)\)
c, Theo pthh ta có: \(n_{NaOH}=2.0,2=0,4\left(mol\right)\)
=>mNaOH=n.M=0,4.(23+16+1)=16(g)
a, PTHH: 2Na + 2H2O ---> 2NaOH + H2
Mol PT: 2 : 2 : 2 : 1
Mol đề bài: 0,2
b,Ta có: nH2= 4,48/22,4=0,2 (mol)
=> nNa= 0,2*2/1=0,2 (mol)
=> mNa= 0,4*22,4=8,96 (g)
c, Ta có: nNaOH=0,2*2/1=0,4 (mol)
=> mNaOH=0,4*40=16 (g)
. \(n_{H_2}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
a. PTHH: \(2Na+2H_2O\rightarrow2NaOH+H_2\)
Theo PT: 2...........2...............2..................1
Theo đề: .................................................0,2
b.
\(\Rightarrow n_{Na}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\Rightarrow m_{Na}=0,4.23=9,2\left(g\right)\)
c. \(\Rightarrow n_{NaOH}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\Rightarrow m_{NaOH}=0,4.40=16\left(g\right)\)
