Sodium metal forms at the cathode
A single replacement reaction could look like this:
2FeCl3 + 3Ba ➡️ 3BaCl2 + 2Fe
In this reaction, the barium is replacing the iron bound to the chlorine.
126 grams of H2O is formed.
Explanation:
Data given:
volume of the gas = 88 Liters
pressure = 720 mm Hg or 0.947 atm
temperature T = 22 Degrees or 295.15 K
R = 0.08021 atm L/mole K
n =?
The formula is used is of ideal gas law to know the number of moles of CH4 undergoing combustion.
PV = nRT
n = 
putting the values in the equation
= 0.947 X 88/ 0.08021 X 295.15
n = 3.5 moles
balanced reaction for combustion of methane
CH4 + O2 ⇒ CO2 + 2H20
1 mole of CH4 undergoes combustion to form 2 moles of water
3.5 moles will give x moles of water
2/1 = x/3.5
x = 7 moles of water (atomic mass of water = 18 gram/mole)
mass = atomic mass x number of moles
mass = 18 x 7
=126 grams of water is formed.
Answer:

Explanation:
Hello there!
In this case, since there is a 2:2 mole ratio between sodium peroxide and water according to the given reaction, it is possible to apply the following stoichiometric setup for the calculation of the required mass of water:

Best regards!
The balanced equation for the above reaction is;
2K + Cl₂ ---> 2KCl
Stoichiomtery of K to KCl is 2:2
Potassium is the limiting reactant which is fully consumed in the reaction. The amount of product formed depends on amount of limits reactant present.
Number of moles of K reacted - 6.75 g/ 39 g/mol = 0.17 mol
Therefore number of KCl moles formed - 0.17 mol
Mass of KCl formed - 0.17 mol x 74.5 g/mol = 12.67 g