4.6克的钠与足量的水反应，计算生成的气体在标准状况下的体积，（标准状况下氢气的密度为0.0899g/L)

Question

如题所述

Answer 1

n(Na)=4.6/23=0.2mol
n(H2)=0.1mol
m(H2)=0.2g
V(H2)=0.2/0.0899=2.2247L

Answer 2

2Na+2H2O=2NaOH+H2
46 2
4.6 x
x=0.2g
V(H2)=0.2g/(0.0899g/L)= 2.22L本回答被网友采纳

Answer 3

2.24L