Question

Sodium hydroxide reacts with carbon dioxide as follows: 2 NaOH(s) + CO2(g) → Na2CO3(s) + H2O(1) If 3.70 mol NaOH and 2.00 mol CO2 are allowed to react, how many moles of the excess reactant remains? O 1.00 mol CO2 0 1.70 NaOH O 0.30 mol NaOH O 0.15 mol CO2

Answer 1

Answer : 0.15 mol of CO₂ will remain as excess reactant

(Answer option 4th will be correct)

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Solution :

The given reaction           :   2NaOH _(s) + CO_{2 (g)} $\rightarrow$ Na₂CO_{3 (s)} + H₂O _(l)

Mole ratio                      :    2 mol         : 1 mol                1 mol        : 1 mol

(ratio of coefficients)

Available moles (given mol) :   3.70mol    : 2 mol                 -             ;    -

From the above mole ratio it is evident that 2 moles of NaOH needs 1 mol of CO₂ therefore the moles of CO₂ needed to react with the avilable 3.70 moles of NaOH will be

                                   

3.70 moles of NaOH x 1 mole of CO2 II 2 moles of NaOH = 1.85 mol CO2

The moles of needed to react with 3.70 moles will be 1.85

The excess of CO₂ remains = moles of CO₂ available - molesof CO₂ needed to react

                                             = 2.00 - 1.85

The excess of CO₂ left           = 0.15 mol of CO₂

Excess of moles of CO₂ remains = 0.15 mol

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From the above calculation it is clear that 0.15 mol of CO₂ remains as excess reactant.Therefore all other answer options are wrong.