Question

The reaction that gives \[C{O_2}\] as one of the product is:
\[A)\;3C + 4HN{O_3}\xrightarrow{\Delta }\]
\[B)\;6NaOH + 2C\xrightarrow{{}}\]
\[C)\;Sn{O_2} + 2C\xrightarrow{{}}\]
\[D)\;Fe2{O_3} + 3C\xrightarrow{{250^\circ - 400^\circ C}}\]

Answer 1

Hint: For solving this type of question, we need to check all the above options present, whether they give $CO_2$ as the product in their reaction or not . By following this step, we would get our final result.So,
\[3C + 4HN{O_3}\xrightarrow{\Delta }3C{O_2} + 4N{O_3} + 2{H_2}O\]
Complete step by step solution:
In this type of problems, we need to at first complete the following reactions or chemical reactions and then balance it so that both the left hand side (L.H.S.) and right hand side (R.H.S.) should have equal moles or number of elements present.
\[3C + 4HN{O_3}\xrightarrow{\Delta }\;?\]
So, \[3C + 4HN{O_3}\xrightarrow{\Delta }C{O_2} + N{O_2} + {H_2}O\]
In question it’s given that ‘3’ Carbon (C); 4 Hydrogen (H), 4 Nitrogen (N) and (3 × 4) = 12 Oxygen (O) is present on the left hand side.
In the similar way 3 carbon, 4 Hydrogen 4 Nitrogen and 12 Oxygen should also be present on the
right hand side. So equation it we get :-
\[3C + 4HN{O_3}\xrightarrow{\Delta }3C{O_2} + 4N{O_2} + 2{H_2}O\]
\[6NaOH + 2C\xrightarrow{{}}\;?\]
So, \[6NaOH + 2C\xrightarrow{{}}{H_2} + Na + N{a_2}C{O_3}\]
In question it’s given that 6Na (sodium), 6 oxygen (O) and 6 Hydrogen (H) and 2 carbon (C) is present on the left hand side. In the similar way, 6 Na, O, H and 2C should also be present on the right hand side. So equating it we get:
\[6NaOH + 2C\xrightarrow{{}}3{H_2} + 2Na + 2N{a_2}C{O_3}\]
 \[Sn{O_2} + 2C\xrightarrow{{}}\;?\]
So, \[Sn{O_2} + 2C\xrightarrow{{}}Sn + 2CO\]
In question it’s given that 1 Sn, 2 Oxygen (O) and 2 Carbon (C) is present on the left hand side.
In the similar way, 1 Sn, 2 Oxygen (O) and 2 Carbon (C) must be present on the right hand side.
So equating it we get :-
\[Sn{O_2} + 2C\xrightarrow{{}}Sn + 2CO\]
\[Fe2{O_3} + 3C\xrightarrow{{250^\circ - 400^\circ C}}\]
So, \[Fe2{O_3} + 3C\xrightarrow{{250^\circ - 400^\circ C}}Fe + CO\]
In question it’s given that 2 Iron (Fe), 3 Oxygen (O) and 3 Carbon (C) is present on the left hand side.
In the similar manner, 2 iron (Fe), 3 Oxygen (O) and 3 Carbon (C) must be present on the right hand side.
So, equating both the sides;
\[F{e_2}{O_3} + 3C\xrightarrow{{}}2Fe + 3CO\]
So, option (a) i.e., chemical reaction:
\[3C + 4HN{O_3}\xrightarrow{\Delta }3C{O_2} + 4N{O_2} + 2{H_2}O\] is the reaction which gives out \[C{O_2}\]

Note:
For solving these chemical equations, we must keep one thing in mind that we must balance the chemical equations on both sides whenever it is asked for the product of the chemical reaction.
\[3C + 4HN{O_3}\xrightarrow{\Delta }3C{O_2} + 4N{O_2} + 2{H_2}O\]
Carbon dioxide is produced whenever an acid reacts with a carbonate. This makes carbon dioxide easy to make in the laboratory. Calcium carbonate and hydrochloric acid are usually used because they are cheap and easy to obtain. Carbon dioxide can be collected over water.