In a reaction, 224 g of iron (Fe) are available that reactcompletely with 96 g of oxygen gas (O2) to form iron trioxide(Fe2O3).

Comments (2) on “In a reaction, 224 g of iron (Fe) are available that reactcompletely with 96 g of oxygen gas (O2) to form iron trioxide(Fe2O3).”

1. hi you are my friend or not

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2. 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.

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