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?

Write the equation and balance it.

Convert what you have in grams to mols.
Convert mols of each reactant to mols of the product. Take the smaller number, which is the limiting reagent. Then convert mols of that reagent to grams water.

To calculate the mass of water produced and determine the limiting reagent in a chemical reaction, we need to follow these steps:

Step 1: Write the balanced chemical equation for the reaction.
The balanced equation for the reaction of hydrogen (H2) and oxygen (O2) to form water (H2O) is:

2H2 + O2 → 2H2O

Step 2: Find the molar masses of the reactants and products.
The molar mass of hydrogen (H2) is 2 g/mol, and the molar mass of oxygen (O2) is 32 g/mol. The molar mass of water (H2O) is 18 g/mol.

Step 3: Convert the given masses of the reactants to moles.
To convert the mass of H2 (24.0 g) to moles, divide the mass by its molar mass:
24.0 g H2 ÷ 2 g/mol = 12.0 mol H2

To convert the mass of O2 (160.0 g) to moles, divide the mass by its molar mass:
160.0 g O2 ÷ 32 g/mol = 5.0 mol O2

Step 4: Determine the mole ratio of the reactants.
From the balanced equation, the mole ratio of H2 to O2 is 2:1.

Step 5: Identify the limiting reagent.
The limiting reagent is the reactant that will be consumed completely and limits the amount of the product that can be formed. To determine the limiting reagent, we compare the number of moles of each reactant with respect to the mole ratio.

The mole ratio of H2 to O2 is 2:1. Since we have 12.0 moles of H2 and 5.0 moles of O2, we can see that the O2 is in excess because we would only need half as many moles of O2 (2.0 mol) to react with all of the H2. Therefore, O2 is the excess reagent and H2 is the limiting reagent.

Step 6: Calculate the mass of water produced.
Using the mole ratio from the balanced equation, we know that 2 moles of H2 will produce 2 moles of H2O. Therefore, the mole ratio of H2 to H2O is 2:2.

Since we have 12.0 moles of H2, we know that we will produce 12.0 moles of H2O.

To convert moles of H2O to grams, multiply the moles by the molar mass of H2O:
12.0 mol H2O × 18 g/mol = 216 g H2O

Therefore, the mass of water produced from the reaction of 24.0 g of H2 and 160.0 g of O2 is 216 grams. The limiting reagent is H2.