If the reaction of 0.500 grams of aluminum metal with excess aqueous HCl releases 9.72 kJ of heat, what is the ΔH value for the reaction? 2 Al(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2(g)
All I have is:
.500gAl*(1 mol Al/26.98gAl)
Please and thank you so much!
I would do it this way. 0.5 g produces 9.72 kJ so how much will you get from 2*26.98?
9.72 kJ x (2*26.98)/0.5) = ?
To find the ΔH (enthalpy change) value for the reaction, you need to calculate the moles of aluminum reacted.
Given: .500g Al
Using the molar mass of aluminum (Al) from the periodic table (26.98 g/mol), you can convert grams of Al to moles of Al:
.500g Al * (1 mol Al / 26.98g Al) = 0.0185 mol Al
Next, since the reaction is balanced in terms of mole ratio, we can use the coefficients of the balanced equation to relate the moles of Al to the heat released.
From the balanced equation, we see that 2 moles of Al produce 9.72 kJ of heat.
Therefore, the ΔH value can be calculated using the following equation:
ΔH = (9.72 kJ / 2 mol) * (1 mol / 0.0185 mol Al)
ΔH = 526.49 kJ/mol
So, the ΔH value for the reaction is 526.49 kJ/mol.
Please note that this calculation assumes that all other conditions (e.g., temperature, pressure) are constant and that the reaction occurs under standard conditions.