The volume ratio of hydrogen released by the reaction of the same amount of sodium, magnesium and aluminum with enough hydrochloric acid is () The volume ratio of hydrogen released by the reaction of sodium magnesium aluminum with enough hydrochloric acid is ()

The volume ratio of hydrogen released by the reaction of the same amount of sodium, magnesium and aluminum with enough hydrochloric acid is () The volume ratio of hydrogen released by the reaction of sodium magnesium aluminum with enough hydrochloric acid is ()

[1] The volume ratio of hydrogen released by the reaction of the same amount of sodium, magnesium and aluminum with enough hydrochloric acid is: (1:2:3)
[2] The volume ratio of hydrogen released by the reaction of sodium magnesium aluminum with enough hydrochloric acid of the same mass
(1 / 23: 1 / 12: 1 / 9)

What is the volume ratio of hydrogen produced by the reaction of sodium magnesium aluminum with enough dilute hydrochloric acid under the same conditions
2. What is the volume ratio of sodium magnesium aluminum to sufficient hydrochloric acid to produce hydrogen under the same conditions? 3. What is the mass ratio of sodium magnesium aluminum required to produce the same volume of hydrogen at the same temperature and pressure? 4. What is the mass ratio of sodium magnesium aluminum required to produce the same volume of hydrogen at the same temperature and pressure

1. Write the ion equation first
Then assume that it's all 1mol metal, and it's intuitive to know the volume ratio of hydrogen
2. Assuming that the mass is m, we can calculate the amount of metal. In the equation, we also know clearly the volume ratio of hydrogen
3. Suppose that 1mol hydrogen is generated, calculate the amount of metal material needed, and then compare
4. Convert the amount of metal in 3 into mass, and then compare
Do it yourself

If the volume ratio of hydrogen released is 1:2:3, the mass ratio of sodium, magnesium and aluminum is -
Then the mass ratio is

When the volume ratio of H2 is 1:2:3, the mass ratio of H2 is 1:2:3
For each electron lost, an H is formed
Then the mass ratio of H2 is the mass ratio of the metal to lose electrons
One Na loses one electron, two mg and three al
The ratio of the amount of substance is 1:1:1
The mass ratio is 23:24:27