Na2CO3 reacts with dil. HCl to produce NaCl, H2O and CO2. If 21.2 g of pure Na2CO3 are added in a solution containing 21.9g HCl , a. Find the limiting reagent. (2) b. Calculate the number of moles of excess reagent left over.(2) c. Calculate the number of molecules of H2O formed.(1) d. Calculate volume of CO2 gas produced at 270C and 760mm Hg pressure.(2) e. Write significance of limiting reagent​

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Explanation:

Equation of the reaction;

Na2CO3(aq) + 2HCl(aq) > 2NaCl(aq) + H2O(l) + CO2(g)

Number of moles of Na2CO3 = 21.2g/106g/mol = 0.2 moles Na2CO3

Number of moles of HCl = 21.9g/36.5g/mol = 0.6 moles of HCl

1 mole of Na2CO3 reacts with 2 moles of HCl

0.2 moles of Na2CO3 reacts with 0.2 × 2/1 = 0.4 moles of HCl

Hence Na2CO3 is the limiting reactant

Since there is 0.6 moles of HCl present, the number of moles of excess reagent=

0.6 moles - 0.4 moles = 0.2 moles of HCl

1 mole of Na2CO3 forms 1 mole of water

0.2 moles of Na2CO3 forms 0.2 moles of water

Number of molecules of water formed = 0.2 moles × 6.02 × 10^23 = 1.2 × 10^23 molecules of water

1 mole of Na2CO3 yields 1 mole of CO2

0.2 moles of Na2CO3 yields 0.2 moles of CO2

1 mole of CO2 occupies 22.4 L

0.2 moles of CO2 occupies 0.2 × 22.4 = 4.48 L at STP

Hence;

V1=4.48 L

T1 = 273 K

P1= 760 mmHg

T2 = 27°C + 273 = 300 K

P2 = 760 mmHg

V2 =

P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

V2 = P1V1T2/P2T1

V2 = 760 × 4.48 × 300/760 × 273

V2= 4.9 L

The limiting reactant is the reactant that determines the amount of product formed in a reaction. When the limiting reactant is exhausted, the reaction stops.

h let me think

pure water does not conduct electricity very well. however, when certain substances are dissolved in water, the solution does conduct electricity. you can make a simple device that shows how well a solution conducts electricity. this device uses a flashlight bulb to indicate how well the solution conducts electricity. the better the solution conducts electricity, the brighter the bulb will glow.

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