Answer:
formula pottasium Oxalate C2k204
Explanation:
C2k204
Molar mass of copper chloride is 134.45 g/mole, so the mole of 10.5 g copper chloride is 10.5/134.45 = 0.078 mole. Molar mass of aluminum is 27 g/mole, so the mole of 12.4 g aluminium is 12.4/27 = 0.46 mole. The formula of this reaction is as follows, 2Al + 3CuCl2 ⇒ 2AlCl3 + 3Cu. Thus the molar ratio of the reactants is Al:CuCl2 = 2:3. So to react with 0.078 mole of copper chloride, will need 0.078 x 2/3 = 0.052 moles of aluminum which is less than the given amount (0.46mole). Therefore, copper chloride is the limiting reactant.
Answer: 0.0746 grams of hydrogen
Explanation:
To calculate the mass of hydrogen :
(Vapor pressure of water is 23.78 mmHg at 25 ∘C )
According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number 
of particles.
According to the ideal gas equation:
P = Total Pressure = 752 mm Hg
pressure of hydrogen = Total Pressure - Vapor pressure of water = (752 -23.78 )mm Hg = 728.22 mm Hg = 0.958 atm (760mmHg=1atm)
V= Volume of the gas = 0.953 L
T= Temperature of the gas = 25°C = 298 K
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas= ?
Mass of 
Thus 0.0746 grams of hydrogen is collected.
Mass of H₂ produced = 12 g
<h3>Further explanation</h3>
Given
32 grams of CH₄
Required
mass of H₂
Solution
Reaction
CH₄ + H₂O (+ heat) → CO + 3H₂
mol CH₄(MW= 16 g/mol) :
= mass : MW
= 32 g : 16 g/mol
= 2
From the equation, mol ratio CH₄ : H₂ = 1 : 3, so mol H₂ :
= 3/1 x mol CH₄
= 3/1 x 2
= 6
Mass H₂(MW = 2 g/mol) :
= mol x MW
= 6 x 2
= 12 g
