Calculate the amount of heat transferred when 2.06 g of Mg(s) reacts at constant pressure.
2 Mg(s) + O2(g) �¨ 2 MgO(s) ƒ¢H = -1204 kJ
I don't understand the symbols.
1o2.01KJ
To calculate the amount of heat transferred during a reaction, we need to use the equation:
q = n * ∆H
where:
q: amount of heat transferred (in joules)
n: number of moles of the substance undergoing the reaction
∆H: enthalpy change of the reaction (in joules per mole)
In this case, we want to calculate the amount of heat transferred when 2.06 g of Mg(s) reacts. Let's calculate the number of moles of Mg(s) first:
molar mass of Mg = 24.31 g/mol
number of moles = mass / molar mass
number of moles of Mg = 2.06 g / 24.31 g/mol
Next, we'll use the stoichiometry of the balanced equation to determine the number of moles of MgO(s) produced. According to the equation:
2 moles of Mg(s) reacts to form 2 moles of MgO(s)
So, the number of moles of MgO(s) will be the same as the number of moles of Mg(s).
Now that we have the number of moles of Mg(s), let's substitute the values into the formula:
q = n * ∆H
q = (2.06 g / 24.31 g/mol) * -1204 kJ/mol
Note: We need to convert kJ to joules for consistency.
1 kJ = 1000 J
q = (2.06 / 24.31) * -1204 * 1000 J
Now we can calculate the value of q.