The thermite reaction is a very exothermic reaction; it has been used to produce liquid iron for welding. A mixture of 2 mol of powdered aluminum metal and 1 mol of iron(III) oxide yields liquid iron and solid aluminum oxide. How many grams of the mixture are needed to produce 313 kJ of heat?
[delta H is -825.5 KJ/mol]
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To determine the number of grams of the mixture needed to produce 313 kJ of heat, we can follow these steps:
Step 1: Determine the number of moles needed to produce 313 kJ of heat.
Using the given value of ΔH = -825.5 kJ/mol, we can calculate the moles of heat produced by using the equation:
ΔH = n × ΔH_molar
where ΔH is the total heat produced, n is the number of moles, and ΔH_molar is the molar enthalpy change.
n = ΔH / ΔH_molar
Plugging in the values, we get:
n = 313 kJ / (-825.5 kJ/mol)
Step 2: Determine the number of moles of mixture needed.
From the balanced equation, we know that 2 mol of aluminum and 1 mol of iron(III) oxide react to produce the desired amount of heat.
Therefore, we need 2n moles of aluminum and n moles of iron(III) oxide. The total number of moles of the mixture will be:
2n + n = 3n
So, the number of moles of the mixture needed to produce 313 kJ of heat is 3n.
Step 3: Convert moles to grams.
Using the molar mass of aluminum (Al) and iron(III) oxide (Fe2O3), we can calculate the mass of the mixture.
Multiply the number of moles of the mixture by the molar mass of the mixture to get the mass:
mass = (3n) × (molar mass of Al + molar mass of Fe2O3)
Step 4: Calculate the mass.
Finally, substitute the value of n from Step 1 into the equation in Step 3 to calculate the mass of the mixture needed to produce 313 kJ of heat.