Balance the following reaction: AL2O3 + HCI ⇒ ALCI3 + H2O

1 answer:

4 0

To balance the following chemical equation, make a tally or a count of each of the atoms on both sides of the reaction, and make sure that those atoms are equal on both the reactant and product side.

AL2O3 + HCl => ALCl3 + H2O

Left side. Right side
AL = 2. AL = 1
O = 3 Cl = 3
H = 1. H = 2
Cl = 1. O = 1

First balance the metal atoms, aluminum, then hydrogen and then oxygen.

Balanced equation :
AL2O3 + 6HCl => 2ALCl3 + 3H2O.

Left side. Right side
AL = 2 AL = 2
O = 3 Cl = 6
H = 6 H = 6
Cl = 6 O = 3.

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N₂O(g) + 3 H₂(g) N₂H4(1) + H₂O(1) AH = -317 kJ/mol

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Answer:

Explanation:

Recall that Δ<em>H</em> is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:

$\displaystyle \Delta H^\circ_{rxn} = \sum \Delta H^\circ_{f} \left(\text{Products}\right) - \sum \Delta H^\circ_{f} \left(\text{Reactants}\right)$

Therefore, from the chemical equation, we have that:

$\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} + \Delta H^\circ_f \text{ H$_2$O} \right] -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}$

Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:

$\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}$