Write a chemical equation for 48g of gasoeous hydrogen and 320g of gaseous oxygen yield

To write a balanced chemical equation for the reaction between gaseous hydrogen (H₂) and gaseous oxygen (O₂), we need to know the products formed. By convention, when hydrogen gas and oxygen gas react, they combine to form water (H₂O). The balanced equation is as follows:

2H₂ + O₂ → 2H₂O

This means that two molecules of hydrogen gas react with one molecule of oxygen gas to produce two molecules of water.

Now, if we want to calculate the yield of water (in grams) formed from the given masses of hydrogen and oxygen, we need to use the concept of stoichiometry. The molar mass of hydrogen (H₂) is 2 g/mol, and the molar mass of oxygen (O₂) is 32 g/mol.

First, we need to determine the number of moles for the given masses:
Number of moles of H₂ = 48 g H₂ / 2 g/mol ≈ 24 mol H₂
Number of moles of O₂ = 320 g O₂ / 32 g/mol ≈ 10 mol O₂

Now we can use the coefficients from the balanced equation to relate the number of moles of reactants to the number of moles of products. In this case, the ratio is 2:1 for H₂ and O₂ to H₂O.

Since we have the limiting reactant (oxygen) with 10 moles, we can calculate the theoretical yield of water using the stoichiometry:

10 mol O₂ × (2 mol H₂O / 1 mol O₂) × (18 g H₂O / 1 mol H₂O) = 360 g H₂O

Therefore, the theoretical yield of water in this reaction is 360 grams.