limitingreactant

# limitingreactant - Questions dealing with Limiting Reactant...

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Questions dealing with: Limiting Reactant If 6 g of hydrogen and 70 g of oxygen react to form water: (a) What is the limiting reactant? (b) How much water can be produced? (c) How many grams of the excess reactant will be left over? First step is to write a balanced equation. 2 H 2 + O 2 2 H 2 O Second, work a stoichiometric problem (which means you will have to use the balanced equation to do the problem} with both of the reactants by themselves. Let's do the problem using 6g of hydrogen first: 6 g H 2 X 1 mole H 2 X 2 moles H 2 O * X 18 g H 2 O = 54 g of H 2 O 1 2 g H 2 2 moles H 2 1 mole H 2 O * The 2 and 2 in this step come from the coefficients in the balanced equation. So, starting with 6 g of H 2 we can make 54 g of water. Now let's so the problem using 70 g of O 2 : 70g O 2 X 1mole O 2 X 2 moles of H 2 O X 18g H 2 O = 78.8 g H 2 O 1 32g O 2 1 mole O 2 1 mole H 2 O So, starting with 70 g of O 2 we can make 78.8 g of H 2 O. Since the hydrogen runs out after making 54 g of water, then 54 g is the amount of water

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limitingreactant - Questions dealing with Limiting Reactant...

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