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Question

# Complete the following equation and balance it:${\mathrm{CO}}_{2}+\mathrm{NaOH}\to$

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Solution

## The reaction can be solved in two steps:Complete the reaction by writing the products.Balance the equation.1.Complete the reaction by writing the products: The reaction of Carbon dioxide with Sodium hydroxide gives Sodium carbonate and water as products.${\mathrm{CO}}_{2}+\mathrm{NaOH}\to {\mathrm{Na}}_{2}{\mathrm{CO}}_{3}+{\mathrm{H}}_{2}\mathrm{O}\phantom{\rule{0ex}{0ex}}\left(\mathrm{Carbon}\mathrm{dioxide}\right)\left(\mathrm{Sodium}\mathrm{hydroxide}\right)\left(\mathrm{Sodium}\mathrm{carbonate}\right)\left(\mathrm{Water}\right)$2. Balancing of the reaction by hit and trial method:A chemical equation is said to be balanced when atoms of every element in the chemical equation are equal in number on both (reactant and product) sides. To balance the chemical equation, begin by multiplying the element or compound which has less number of atoms with a number that would make the number equal on both sides.${\mathrm{CO}}_{2}+\mathrm{NaOH}\to {\mathrm{Na}}_{2}{\mathrm{CO}}_{3}+{\mathrm{H}}_{2}\mathrm{O}\phantom{\rule{0ex}{0ex}}\left(\mathrm{Carbon}\mathrm{dioxide}\right)\left(\mathrm{Sodium}\mathrm{hydroxide}\right)\left(\mathrm{Sodium}\mathrm{carbonate}\right)\left(\mathrm{Water}\right)$Left hand side (LHS) is the reactant side and Right hand side (RHS) is the product side.Type of atomsNumber of atoms on L.H.S.Number of atoms on R.H.S.Carbon (C)11Oxygen (O)34Sodium (Na)12Hydrogen (H)12Multiply Sodium hydroxide (NaOH) on LHS with 2 as the number of atoms of Sodium are less on LHS.${\mathrm{CO}}_{2}+2\mathrm{NaOH}\to {\mathrm{Na}}_{2}{\mathrm{CO}}_{3}+{\mathrm{H}}_{2}\mathrm{O}\phantom{\rule{0ex}{0ex}}\left(\mathrm{Carbon}\mathrm{dioxide}\right)\left(\mathrm{Sodium}\mathrm{hydroxide}\right)\left(\mathrm{Sodium}\mathrm{carbonate}\right)\left(\mathrm{Water}\right)$Type of atomsNumber of atoms on L.H.S.Number of atoms on R.H.S.Carbon (C)11Oxygen (O)44Sodium (Na)22Hydrogen (H)22As the number of all the atoms is equal on both sides, the chemical equation is now balanced.So, the overall complete and balanced chemical equation is:${\mathrm{CO}}_{2}\left(\mathrm{g}\right)+2\mathrm{NaOH}\left(\mathrm{aq}\right)\to {\mathrm{Na}}_{2}{\mathrm{CO}}_{3}\left(\mathrm{aq}\right)+{\mathrm{H}}_{2}\mathrm{O}\left(\mathrm{l}\right)\phantom{\rule{0ex}{0ex}}\left(\mathrm{Carbon}\mathrm{dioxide}\right)\left(\mathrm{Sodium}\mathrm{hydroxide}\right)\left(\mathrm{Sodium}\mathrm{carbonate}\right)\left(\mathrm{Water}\right)$

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