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How much energy is produced when 93.5 grams of oxygen react with 13.2 grams of hydrogen in the following reaction:

2H2+o2-->2 h2o triangleH=-572kJ

User Bunkar
by
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1 Answer

1 vote

Answer:

-1670.24 kJ

QUICK Explanation:

ΔH * moles of limiting reactant = Energy produced

-572 kJ/mol * 2.92 moles of O2 = -1670.24 kJ

LONGER EXPLANATION :

2 H2 + O2 -> 2 H2O

ΔH * moles of limiting reactant = Energy produced

enthalpy change or heat of reaction formula

1. Calculate the number of moles of oxygen (O2):

Number of moles = mass / molar mass

Number of moles of O2 = 93.5 g / 32.00 g/mol

2.92 mol O2

2. Calculate the number of moles of hydrogen (H2):

Number of moles = mass / molar mass

Number of moles of H2 = 13.2 g / 2.02 g/mol

6.53 mol H2

3. Determine the limiting reactant:

According to the balanced equation,

2 moles of H2 react with 1 mole of O2.

Calculate the moles of O2 based on the moles of H2:

(6.53 mol H2) / (2 mol H2/O2) = 3.27 mol O2

we need 3.27 mol O2 to react with the available H2

BUT only have 2.92 mol of O2 available

O2 is the limiting reactant

4.

Calculate the heat given off by assuming the complete consumption of the limiting reagent

calculate the amount of energy produced using the given enthalpy change (ΔH):

Energy produced = ΔH * moles of limiting reactant

Energy produced = ΔH * moles of O2 reacted

Calculate the energy produced using the given enthalpy change (ΔH):

Energy produced = ΔH * moles of O2 reacted

= -572 kJ/mol * 2.92 mol

≈ -1670.24 kJ

Therefore, approximately -1670.24 kJ of energy is produced when 93.5 grams of oxygen react with 13.2 grams of hydrogen in the given reaction. Note that the negative sign indicates that the reaction is exothermic (energy is released).

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User Miguel Vazq
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