Question

The reaction of zinc with hydrochloric acid is represented by the reaction

Zn(s) + 2 HCl(aq) ↔ ZnCl_{2}(aq) + H_{2}(g) --
ΔH = -152.5 kJ

How many grams of zinc reacted with an excess of HCl (100.0 mL) if the temperature of the calorimeter increased from 25.0°C to 31.83 °C? Assume the heat capacity of the solution is the same as pure water (4.184 J/g*°C), the density of the solution is 1.00 g/mL and there is no loss of heat to the surroundings.

Answer #1

Solution"- volume of solution would be same as the volume of HCl that is 100.0 mL

density of solution = 1.00 g/mL

mass of solution = 100.0 mL x 1.00g/mL = 100.0 g

delta T = 31.83 - 25.0 = 6.83 ^{0}C

q = m c deltaT

q = 100.0 x 4.184 x 6.83

q = 2857.672 J = 2.86 kJ

From balanced equation, 152.5 kJ of heat is released when 1 mol of Zn is used. How many moles of Zn are used when 2.86 kJ of heat is released.

2.86 kJ x 1mol/152.5 kJ = 0.0188 mol

We would convert these moles to grams on multiplying the moles with atomic mass of Zn.

0.0188 mol x 65.38g/1mol = **1.23 g**

**So, 1.23 g of Zn were reacted.**

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