the reaction of methane and water is one way to prepare hydrogen for use as a fuel:
methanw react with water to from carbonmonoxide and hydrogen
if you begin with 995g of methane and 2510g of water,
1.which reactant is the limiting reactant?
2.what is the maximum mass of hydrogen that can be prepared?
3.what mass of excess reactant remains when the reaction is completed?
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To determine the limiting reactant, maximum mass of hydrogen, and the mass of excess reactant remaining, we need to follow a series of steps:
Step 1: Write and balance the chemical equation for the reaction:
CH₄ (methane) + 2H₂O (water) → CO (carbon monoxide) + 4H₂ (hydrogen)
Step 2: Convert the given masses of reactants to moles.
Molar mass of CH₄ (methane) = 12.01 g/mol + 4(1.01 g/mol) = 16.05 g/mol
Moles of methane = mass of methane / molar mass of methane = 995 g / 16.05 g/mol = 61.97 mol
Molar mass of H₂O (water) = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
Moles of water = mass of water / molar mass of water = 2510 g / 18.02 g/mol = 139.30 mol
Step 3: Determine the stoichiometric ratio between methane and water to find the limiting reactant.
From the balanced equation, we see that one mole of methane reacts with two moles of water.
Therefore, the moles of water needed to react with the given moles of methane are twice the moles of methane.
Required moles of water = 2 * moles of methane = 2 * 61.97 mol = 123.94 mol
Step 4: Compare the moles of the actual reactants with the required moles to identify the limiting reactant.
Considering the amount of water provided (139.30 mol) is greater than the required amount (123.94 mol), water is in excess. Therefore, the limiting reactant is methane.
Step 5: Calculate the maximum mass of hydrogen that can be produced.
From the balanced equation, we see that for every one mole of methane, four moles of hydrogen are produced.
Moles of hydrogen = 4 * moles of methane = 4 * 61.97 mol = 247.88 mol
Mass of hydrogen = moles of hydrogen × molar mass of hydrogen
Molar mass of hydrogen = 2(1.01 g/mol) = 2.02 g/mol
Maximum mass of hydrogen = 247.88 mol × 2.02 g/mol = 501.10 g
Step 6: Calculate the mass of excess reactant remaining.
Mass of excess water = (moles of water - required moles of water) × molar mass of water
= (139.30 mol - 123.94 mol) × 18.02 g/mol = 277.97 g
Therefore:
1. The limiting reactant is methane.
2. The maximum mass of hydrogen that can be prepared is 501.10 g.
3. The mass of excess reactant remaining when the reaction is completed is 277.97 g of water.