5) How much heat can be produced from a reaction mixture of 50 gof iron (III) oxide and 25 g of aluminium in the thermite reaction: Fe_(2)O_(3)(s)+2Al(s)arrow Al_(2)O_(3)(s)+2Fe(s);Delta H=-851.5kJ/m

To calculate the amount of heat produced in the thermite reaction, we need to determine the limiting reactant and then use the given enthalpy change to find the heat released.<br /><br />First, let's calculate the moles of iron(III) oxide (Fe2O3) and aluminum (Al):<br /><br />Moles of Fe2O3 = mass / molar mass = 50 g / 159.69 g/mol = 0.314 mol<br />Moles of Al = mass / molar mass = 25 g / 26.98 g/mol = 0.923 mol<br /><br />Next, we need to determine the stoichiometric ratio between Fe2O3 and Al in the balanced chemical equation:<br /><br />Fe2O3(s) + 2Al(s) → Al2O3(s) + 2Fe(s)<br /><br />From the balanced equation, we can see that 1 mole of Fe2O3 reacts with 2 moles of Al. Therefore, the stoichiometric ratio is 1:2.<br /><br />Now, let's compare the moles of Fe2O3 and Al to the stoichiometric ratio:<br /><br />Moles of Fe2O3 / 1 = 0.314 mol<br />Moles of Al / 2 = 0.923 mol / 2 = 0.462 mol<br /><br />Since the moles of Al are greater than the moles of Fe2O3, Al is the excess reactant, and Fe2O3 is the limiting reactant.<br /><br />Now, we can calculate the heat produced using the given enthalpy change:<br /><br />Heat produced = Moles of limiting reactant × Enthalpy change<br />Heat produced = 0.314 mol × (-851.5 kJ/mol)<br />Heat produced = -268.6 kJ<br /><br />Therefore, the reaction mixture can produce approximately 268.6 kJ of heat.