For the reaction Al (s) + 3 HCl (aq) à AlCl3 (s) + ₃/₂ H2 (g), it was observed
that when 4.00 g of aluminum reacted with excess HCl, exactly 30.0 kJ of heat
were released. The DHo for this reaction is
a) – 91.3 kJ
b) – 120 kJ
c) – 202 kJ
d) – 7.50 kJ
no idea please leave helpful steps thanks
The problem can be restated as follows: 30 kJ heat released when 4.00 g Al reacts; how much heat released when 1 mol (26.98 g) Al reacts. That will be
30 x (26.98/4.00) = ?
To determine the DHo (standard enthalpy change) for the given reaction, we need to use the formula:
DHo = q / n
where:
- DHo is the standard enthalpy change
- q is the heat released or absorbed in the reaction
- n is the number of moles involved in the reaction
In this case, we are given the heat released (q) as 30.0 kJ and the mass of aluminum (Al) as 4.00 g. To find the number of moles of aluminum (n), we need to divide the mass by the molar mass of aluminum.
Step 1: Calculate the number of moles of aluminum (Al)
The molar mass of aluminum is 26.98 g/mol. To find the number of moles, use the following formula:
moles = mass / molar mass
moles = 4.00 g / 26.98 g/mol
Step 2: Calculate the DHo
Now, we can substitute the given value of q and the calculated value of n into the formula:
DHo = q / n
DHo = 30.0 kJ / (4.00 g / 26.98 g/mol)
DHo = 30.0 kJ / (4.00 g / 26.98 g/mol)
DHo = (30.0 kJ) * (26.98 g/mol) / 4.00 g
DHo = 202.35 kJ
Therefore, the DHo for the reaction is approximately 202 kJ. The correct answer is (c) – 202 kJ.