What is the largest amount of H2O in grams that can be formed when 4.90 g of H2 reacts with 4.90 g of O2, based on the balanced reaction below.
2 H2 + O2 2 H2O
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This is a limiting reagent problem and I work these the long way by solving two stoichiometry problems and taking the smaller value.
Convert 4.90 g H2 to moles. moles = grams/molar mass.
Convert4.90 moles O2 to moles the samw way.
Using the coefficients in the balanced equation, convert moles H2 to moles H2O.
With the same procedure, convert moles O2 to moles H2O.
You will obtain two answers; of course both can't be right. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that number is the limiting reagent.
Then grams H2O = moles H2O x molar mass H2O.
To solve this problem, we need to determine which reactant is the limiting reactant. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed.
First, we'll find out how much H2O is formed by each reactant individually.
For 4.90 g of H2, we can use its molar mass to convert grams to moles:
Molar mass of H2 = 2 g/mol
Number of moles of H2 = 4.90 g / 2 g/mol = 2.45 mol
According to the balanced equation, 2 moles of H2 produce 2 moles of H2O. Therefore, 2.45 moles of H2 will produce 2.45 moles of H2O.
Next, let's calculate the amount of H2O formed from the O2.
For 4.90 g of O2, we can use its molar mass to convert grams to moles:
Molar mass of O2 = 32 g/mol
Number of moles of O2 = 4.90 g / 32 g/mol ≈ 0.153 mol
According to the balanced equation, 1 mole of O2 produces 2 moles of H2O. Therefore, 0.153 moles of O2 will produce 2 * 0.153 = 0.306 moles of H2O.
Now we compare the amounts of H2O produced by each reactant to determine the limiting reactant.
From H2: 2.45 moles of H2O
From O2: 0.306 moles of H2O
Since the balanced equation requires equal amounts of H2 and O2 to produce H2O, we can compare the moles of H2O formed to see which reactant is limiting.
In this case, it is clear that O2 is the limiting reactant because it produces less moles of H2O compared to H2. Therefore, we can determine the amount of H2O formed based on the limiting reactant, O2.
To find the mass of H2O formed, we multiply the number of moles of H2O by its molar mass:
Molar mass of H2O = 18 g/mol
Mass of H2O = 0.306 mol * 18 g/mol ≈ 5.51 g.
So, the largest amount of H2O that can be formed when 4.90 g of H2 reacts with 4.90 g of O2 is approximately 5.51 grams.