How many grams of oxygen are necessary to produce 8.0 g of water? (molar mass of h=1g/mol, molar mass of 0=16g/mol

7.11grams of oxygen are necessary to produce 8.0 grams of water.

Explanation: hydrogen and oxygen combine at certain ratios to produce water. Since we know that 2 moles of hydrogen, basically combines with 1 mole of oxygen to form 1 mole of water.

Let us consider a simple chemical reaction for the production of water:

2H2(g) + O2(g) > 2H2O(l)

2mol: 1 mol : 2 mol

Considering their masses,

•Molar mass of hydrogen= 1g/mol, as 2H2 we have 2(1g/mol×2) = 4g/mol

•Molar mass of oxygen= 16g/mol, as O2 we have 2×16g/mol= 32g/mol

•Molar mass of water, H2O= (2×1g/mol) + 16g/mol= 2g/mol+ 16g/mol= 18g/mol, as 2H20 , we simply mulitply by 2

So that we have 2×18g/mol = 36g/mol

•Since the question only relates the mass of oxygen to the mass of water produced,

•From the equation, 32g/mol of oxygen will produce 36g/mol of water.

• If O2 2H2O

32g/mol O2 produce 36g/mol H2O

Xg of O2 will produce 8g of H2O

Cross multiply

Xg= 32gmol-1 × 8g / 36gmol-1

Xg= 256g/ 36 = 7.11g

Therefore, 7.11grams of oxygen are necessary to produce 8.0 grams of water.