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Chapter 7: Problem 99

The enthalpy of neutralization for the reaction of a strong acid with a strong base is \(-56 \mathrm{kJ} / \mathrm{mol}\) water produced. How much energy will be released when \(200.0 \mathrm{mL}\) of \(0.400 \mathrm{M} \mathrm{HNO}_{3}\) is mixed with \(150.0 \mathrm{mL}\) of \(0.500 \mathrm{M}\) KOH?

Short Answer

Expert verified
When 200.0 mL of 0.400 M HNO₃ is mixed with 150.0 mL of 0.500 M KOH, the limiting reactant is KOH, and 0.075 mol of water is produced. The energy released during the reaction is \(0.075 \mathrm{mol} \times (-56 \mathrm{kJ/mol}) = -4.2 \mathrm{kJ}\). Thus, 4.2 kJ of energy is released.

Step by step solution

01

Calculate the moles of the reactants

We are given the volume and molarity of both HNO₃ and KOH. We can use the formula n = M × V to find the moles of the reactants. For HNO₃: n(HNO₃) = 0.400 M × 200.0 mL = 0.400 mol/L × 0.200 L = 0.080 mol For KOH: n(KOH) = 0.500 M × 150.0 mL = 0.500 mol/L × 0.150 L = 0.075 mol
02

Determine the limiting reactant

Since the reaction involves the combination of an acid (HNO₃) and a base (KOH), the balanced chemical equation looks like this: HNO₃ + KOH → KNO₃ + H₂O From the balanced equation, we can see that the ratio of moles between HNO₃ and KOH is 1:1. We need to determine which one of the reactants is the limiting reactant by comparing their mole ratio. Since 0.080 mol of HNO₃ reacts with 0.075 mol of KOH (1:1 ratio), HNO₃ is present in excess, and KOH is the limiting reactant.
03

Calculate the moles of water produced

The limiting reactant determines the moles of water produced. In this case, since KOH is the limiting reactant, 0.075 mol of KOH produces 0.075 mol of H₂O.
04

Calculate the energy released

Now that we have determined the number of moles of water produced, we can calculate the energy released. We know that the enthalpy of neutralization is -56 kJ/mol of water. Therefore, we can use the following formula: Energy released = moles of water × enthalpy of neutralization per mole of water Energy released = 0.075 mol × (-56 kJ/mol) = -4.2 kJ Since the energy released is negative, this tells us that 4.2 kJ of energy is released during the reaction.

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Hydrogen gives off \(120 .\) J/g of energy when burned in oxygen, and methane gives off \(50 .\) J/g under the same circumstances. If a mixture of 5.0 g hydrogen and \(10 .\) g methane is burned, and the heat released is transferred to \(50.0 \mathrm{g}\) water at \(25.0^{\circ} \mathrm{C},\) what final temperature will be reached by the water?

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