How many grams of water are produced from the reaction of 1.26 mol of oxygen according to this equation? 2H2(g) + O2(g) �¨ 2H2O(g)

From the equation you can see that each mole of O2 produced 2 moles of water

so 1.26 mole of O2 produces 2.52 mole water

Then use mass = moles x molar mass

to calculate the mass of water produced.

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To determine the number of grams of water produced from the reaction of 1.26 mol of oxygen, we need to use the stoichiometry of the balanced equation.

From the balanced equation:
2H2(g) + O2(g) → 2H2O(g)

We can see that the ratio between O2 and H2O is 1:2.

1 mol of O2 reacts to form 2 mol of H2O.

Therefore, for every 1.26 mol of O2, we will have (1.26 mol O2) × (2 mol H2O / 1 mol O2) = 2.52 mol of H2O.

To convert mol to grams, we need to use the molar mass of water, which is approximately 18.015 g/mol.

(2.52 mol H2O) × (18.015 g H2O / 1 mol H2O) ≈ 45.38 g of H2O.

Therefore, approximately 45.38 grams of water are produced from the reaction of 1.26 mol of oxygen.

To find out how many grams of water are produced from the reaction, we first need to determine the mole ratio between oxygen and water based on the balanced equation.

The balanced equation is: 2H2(g) + O2(g) → 2H2O(g)

From the equation, we can see that for every 1 mole of O2, 2 moles of H2O are produced.

Given that we have 1.26 mol of oxygen (O2), we can use the mole ratio to calculate the number of moles of water produced:

1.26 mol O2 × (2 mol H2O / 1 mol O2) = 2.52 mol H2O

Now, to convert moles of water to grams, we need to use the molar mass of water (H2O), which is approximately 18.015 g/mol.

2.52 mol H2O × 18.015 g/mol = 45.3478 g H2O

Rounding to the appropriate number of significant figures, the mass of water produced from the reaction of 1.26 mol of oxygen is approximately 45.35 grams.