Question

3 seperate questions i have been having trouble with

Suppose you are investigating the reaction: M(s) + 2 HCl(aq) - MC12(aq) + H2(g). You weigh out a 0.243 gram piece of metal and combine it with 63.1 mL of 1.00 M HCl in a coffee-cup calorimeter. If the molar mass of the metal is 46.62 g/mol, and you measure that the reaction absorbed 105 J of heat, what is the enthalpy of this reaction in kJ per mole of limiting reactant? Enter your answer numerically to three significant figures in units of kJ/mol.

Answer 1

1) The moles of the metal are calculated:

n Metal = g / MM = 0.243 g / 46.62 g / mol = 0.0052 mol

Enthalpy change is calculated:

ΔH = - q / n = - 0.105 kJ / 0.0052 mol = - 20.19 kJ / mol

2) The heat transferred is calculated:

q = m * cp * ΔT = 100 * 4,184 * (34,037 - 80,000) = - 19230 J

The molar enthalpy is calculated:

ΔH = 19.23 kJ * 78.398 g / mol / 7.564 g = 199.3 kJ / mol

3) The reaction enthalpy is calculated:

ΔH = ΔHf products - ΔHf reagents = (-497,884) + (2 * 238,181) - (2 * 261,259) - (-484,147) = - 59,893 kJ / mol

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