False it is actually called a neutralization reaction.
Answer:
the volume occupied by 3.0 g of the gas is 16.8 L.
Explanation:
Given;
initial reacting mass of the helium gas, m₁ = 4.0 g
volume occupied by the helium gas, V = 22.4 L
pressure of the gas, P = 1 .0 atm
temperature of the gas, T = 0⁰C = 273 K
atomic mass of helium gas, M = 4.0 g/mol
initial number of moles of the gas is calculated as follows;

The number of moles of the gas when the reacting mass is 3.0 g;
m₂ = 3.0 g

The volume of the gas at 0.75 mol is determined using ideal gas law;
PV = nRT

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.
The closer to the top the metal is in the list, the more active the metal is and the stronger a reducing agent the metal is. When two different metals are involved in a redox reaction, the metal higher in the list will be oxidized and give up electrons that will reduce the cation of the less active metal.
Because water vapor, uncontained, will diffuse out into the atmosphere and is not close enough to condense back into the liquid water, so no equilib. can be established.