Question

cho mau na vao nuoc thay cho 11,2 (1)khi bay len (o dktc) tinh na

a 2,3 g

b1,15g

c0,5g

d2,4g

Answer 1

\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH: 2Na + 2H₂O --> 2NaOH + H₂

              1<------------------------0,5

=> m_Na = 1.23 = 23 (g)

Answer 2

\(pthh:2Na+2H_2O--->2NaOH+H_2\)

Ta có: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Theo pt: \(n_{Na}=2.n_{H_2}=2.0,5=1\left(mol\right)\)

\(\Rightarrow m_{Na}=1.23=23\left(g\right)\)