Question

Suppose you are investigating the reaction: M(s) + 2 HCl(aq) → MCl₂(aq) + H₂(g).

You weigh out a 0.245 gram piece of metal and combine it with 58.7 mL of 1.00 M HCl in a coffee-cup calorimeter. If the molar mass of the metal is 42.75 g/mol, and you measure that the reaction absorbed 160 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

Comment in case of any doubt.

M(S) + Qtuca) Mel (as) + H₂18) : (42.75 at = 5.73x10 ml C moles of the metal = 0.245g -3 P2.75 g/mol) = 0.00573 mol moles of He required for 0.00573 mal metal=240.00573=0.01146 and moles of He added = 1 mol , 0.0587 L= 0.0587 mal Since moles of metal M added are less added Hee are in excess of required 0.01146mal, so Metal is limiting reactant. Thus, moles of reaction occurring = 5.73x50 nal; enthalpy of reaction, Alan = + 160 I 5.73%253 mal = 27923.2 J molti = 127.9 KJ matt