Calculate the mass of water produced from the reaction of 24.0 g of H2 and 160.0 g of O2. What is the limiting reagent?
180g H2O ; Oxygen
To find the mass of water produced and determine the limiting reagent, we first need to write the balanced chemical equation for the reaction between H2 and O2 to form H2O:
2H2 + O2 --> 2H2O
The ratio from the balanced equation tells us that for every 2 moles of H2 consumed, 2 moles of H2O are produced.
Step 1: Calculate the number of moles of H2 and O2:
- Moles of H2 = mass of H2 / molar mass of H2
= 24.0 g / 2.02 g/mol
= 11.88 mol
- Moles of O2 = mass of O2 / molar mass of O2
= 160.0 g / 32.00 g/mol
= 5.00 mol
Step 2: Determine the limiting reagent:
To find the limiting reagent, compare the mole ratios of H2 and O2 in the chemical equation with the actual moles we have.
From the balanced equation, we can see that for every 2 moles of H2, 1 mole of O2 is needed. Therefore, the ratio of moles of H2 to moles of O2 should be 2:1.
The actual ratio of moles of H2 to moles of O2 is:
11.88 mol H2 / 5.00 mol O2 = 2.376
Since the ratio is greater than 2, it means we have an excess of H2 and O2 is the limiting reagent.
Step 3: Calculate the mass of water produced:
The mole ratio from the balanced equation tells us that 2 moles of H2O are produced for every 2 moles of H2 consumed. Therefore, the ratio of H2O produced to H2 consumed is 1:1.
Since O2 is the limiting reagent, it determines the amount of H2 consumed and the amount of H2O produced. Therefore, the moles of H2O produced will be equal to the moles of O2.
Moles of H2O = moles of O2 = 5.00 mol
To calculate the mass of water produced, we use the equation:
Mass of H2O = moles of H2O * molar mass of H2O
The molar mass of H2O = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
Mass of H2O = 5.00 mol * 18.02 g/mol = 90.1 g
Therefore, the mass of water produced from the reaction is 90.1 grams. The limiting reagent in this reaction is O2.
To calculate the mass of water produced from the reaction, you first need to determine the limiting reagent. The limiting reagent is the reactant that gets completely consumed and determines the amount of product that can be formed.
To find the limiting reagent, you can compare the moles of each reactant with the stoichiometry of the balanced chemical equation.
The balanced equation for the reaction between H2 and O2 to produce water is:
2H2 + O2 -> 2H2O
Step 1: Convert the given masses of H2 and O2 to moles.
Molar mass of H2 = 2.02 g/mol
Molar mass of O2 = 32.00 g/mol
Number of moles of H2 = 24.0 g / 2.02 g/mol = 11.88 mol
Number of moles of O2 = 160.0 g / 32.00 g/mol = 5.00 mol
Step 2: Determine the mole ratio between H2 and O2 based on the balanced equation.
From the balanced equation, the mole ratio between H2 and O2 is 2:1. This means that for every 2 moles of H2, 1 mole of O2 is required to react.
Step 3: Compare the moles of H2 and O2 to determine the limiting reagent.
Since the mole ratio of H2 to O2 is 2:1, we can see that 1 mole of O2 is not enough to react with 11.88 moles of H2. Therefore, the limiting reagent is O2.
Step 4: Calculate the mass of water produced.
Since O2 is the limiting reagent, it will be completely consumed in the reaction. Thus, the number of moles of water produced will be equal to the number of moles of O2.
Number of moles of water produced = 5.00 mol
Now, we need to calculate the mass of water produced:
Molar mass of water (H2O) = 18.02 g/mol
Mass of water produced = Number of moles of water produced * Molar mass of water
Mass of water produced = 5.00 mol * 18.02 g/mol = 90.10 g
Therefore, the mass of water produced from the reaction is 90.10 grams, and the limiting reagent is O2.
1. Write the equation and balance it.
2. Convert 24.0 g H2 and 160 g O2 to moles. moles = grams/molar mass.
3a. Using the coefficients in the balanced equation, convert moles H2 to moles H2O.
3b. Do the same for moles O2 to moles H2O.
3c. You have two different answers for moles of the product. Obviously, one of them is correct and one is not. The correct one, in limiting reagent problems. is ALWAYS the smaller one and the reactant giving that number is the limiting reagent.
4. Using the smaller value in 3c, convert to grams. grams = moles x molar mass.
Post your work if you get stuck.