How many mL of 0.098 M HCl is required to completely react with 30.0 mL of 0.150 M K2CO3 solution according to the equation below? What is the molarity of KCl solution formed? 2HCl(aq) + K2CO3(aq) ⟶ 2KCl(aq) + H2O(l) + CO2(g)

It is given that 30.0 mL of 0.150 M K2CO3 is reacting completely with 0.098 M of HCl and the reaction is shown below, 2HCl(aq) + K2CO3(aq) ⟶ 2KCl(aq) + H2O(l) + CO2(g)To calculate the moles of K2CO3 present, no. of moles = molarity × volume in litresmoles of K2CO3= 0.150 M × 0.030 L = 0.0045 molesTo calculate the moles of HCl present with the help of the balanced equation, In the balanced reaction as 1 mole of K2CO3 is reacting with 2 moles of HCl,therefore, 0.0045 moles of K2CO3 will react with 2×0.0045 = 0.009 moles of HCl. To calculate the volume of HCl needed to completely react with K2CO3, volume of HCl in litres = no. of molesmolarity= 0.009 moles0.098 M= 0.0918 L= 91.8 mLTo calculate the molarity of the KCl solution formed, In the balanced reaction as 2 moles of HCl is giving 2 moles of KCl,Therefore, 0.009 moles of HCl will give 0.009 moles of KCl,molarity = no. of molesvolume of solution in litres= 0.009 moles(0.030 + 0.0918) L= 0.0738 M KCl

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Chemistry

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ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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How many mL of 0.098 M HCl is required to completely react with 30.0 mL of 0.150 M K₂CO₃ solution according to the equation below? What is the molarity of KCl solution formed?

2HCl(aq) + K₂CO₃(aq) ⟶ 2KCl(aq) + H₂O(l) + CO₂(g)

Expert Solution

Step 1

It is given that 30.0 mL of 0.150 M K₂CO₃ is reacting completely with 0.098 M of HCl and the reaction is shown below,

$2 {HCl}_{(aq)} + K_{2} {CO}_{3 (aq)} ⟶ 2 {KCl}_{(aq)} + H_{2} O_{(l)} + {CO}_{2 (g)}$

Step 2

To calculate the moles of K₂CO₃ present,

$no . of moles = molarity \times volume in litres moles of K_{2} {CO}_{3} = 0.150 M \times 0.030 L = 0.0045 moles$

Step 3

To calculate the moles of HCl present with the help of the balanced equation,

$In the balanced reaction as 1 mole of K_{2} {CO}_{3} is reacting with 2 moles of HCl, therefore, 0.0045 moles of K_{2} {CO}_{3} will react with 2 \times 0.0045 = 0.009 moles of HCl .$

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