Question

Calculate the mass of water produced from the reaction of 2.00 grams of hydrogen and 4.00 grams of oxygen.​

Answer 1

To calculate the mass of water produced from the reaction of 2.00 grams of hydrogen and 4.00 grams of oxygen, we can use the mole ratio in the balanced chemical equation. For every 2 moles of hydrogen, we get 2 moles of water. Therefore, if we have 2.00 grams of hydrogen, that corresponds to 1 mole of hydrogen. The mass of water produced is 36.04 grams.

Step-by-step explanation:

To calculate the mass of water produced from the reaction of 2.00 grams of hydrogen and 4.00 grams of oxygen, we first need to determine the mole ratio between hydrogen and water in the balanced chemical equation: 2 H₂(g) + O₂(g) → 2 H₂O(g).

From the equation, we can see that for every 2 moles of hydrogen, we get 2 moles of water. Therefore, if we have 2.00 grams of hydrogen, that corresponds to 1 mole of hydrogen.

Since the molar mass of water is 18.02 g/mol, the mass of water produced from the reaction is 2 moles of water x 18.02 g/mol = 36.04 grams.

Answer 2

4.50 g of H₂O

Step-by-step explanation:

The Balance chemical equation is

2 H₂ + O₂ = 2 H₂O

Step 1: Calculate Limiting Reagent

Moles of H₂

Mole = 2 g / 2.02 g/mol = 0.99212 mol

Moles of O₂

Mole = 4 g / 32.00 g/mol = 0.125 mol

Mole ratio of H₂ : O₂ is 2 : 1. Hence, 0.99212 mol of H₂ will need 0.49606 mol of O₂. This means O₂ is limiting reagent.

Step 2: Calculate Moles of H₂O

0.125 mol of O₂ will produce 0.250 mol of H₂O because the mole ratio of O₂ : H₂O is 1 : 2.

Step 3: Calculate mass of H₂O

Mass = 0.250 mol × 18.02 g/mol = 4.50 g of H₂O