a) Determine which of the two reactants (MgCO34) or HCla) is the limiting reagent. b) Determine how much heat (AH ~ kJ) will be produced when: the 1.3443 g of solid magnesium carbonate MgCO34) reacts with the 54.90 mL of 0.656 M hydrochloric acid. HCl«) AH° ra _kJ / (1.3443 g MgCO3) + 54.90 mL of 0.656 M HCla9)
Reaction
MgCO3 (s) + 2HCl(aq) ------> MgCl2 (aq) + H2O (l) + CO2 (g);
From this balanced equation, one mol of MgCO3 reacts with two moles of HCl and forms one mole of MgCl2 one mol of water and one mol of CO2.
Weight of solid MgCO3 = 1.3443 g
Molecular weight of MgCO3 = 84.32 g/mol
Number of moles of MgCO3 = 1.3443 g / 84.32 g/mol
= 0.01594 mol
[HCl] = 0.656 mol/L
Volume of HCl added = 54.90 mL = 0.0549L
Number of moles of HCl added = Concentration(M) * volume in L
= 0.656 mol/L * 0.0549 L
= 0.0360144 mol
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