Question

The following reaction releases 1184.0 kJ of heat:

2 Sr (s)   +   O₂ (g)    ⟶     2 SrO (s)   

What is the heat absorbed/released when 50.0 g of strontium reacts with excess molecular oxygen?

Answer 1

Solution:

2 Sr(s) + O2(g) → 2 SrO(s) ΔHrxn = 1184 kJ

In this case, you can say that when 2 moles of Sr react with 1 mole of oxygen gas, 2 moles of SrO are produced and the enthalpy change of reaction is ΔHrxn = + 1184 kJ.

In other words, when 2 moles of Sr undergo combustion, +1184 kJ" of heat are being absorned, hence the plus sign used for ΔHrxn.

So, you know that the reaction absorb 1184 kJ of heat when two moles of Sr react. Use the compound's molar mass to determine how many moles you get in that 50.0-g sample

moles of Sr = mass of Sr / molar mass = 50 / 87.5 = 0.57 mol Sr

You can now use the known enthalpy change of reaction for when two moles of Sr undergo combustion as a conversion factor to figure out how much heat is being given off when 0.57 moles react

0.57 x (1184 / 2 mol Sr) = 337.44 KJ

So the heat absorbed = 337.44 KJ