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Question: The reaction that gives \[C{O_2}\] as one of the product is: \[A)\;3C + 4HN{O_3}\xrightarrow{\Delt...

The reaction that gives CO2C{O_2} as one of the product is:
A)  3C+4HNO3→ΔA)\;3C + 4HN{O_3}\xrightarrow{\Delta }
B)  6NaOH+2C→B)\;6NaOH + 2C\xrightarrow{{}}
C)  SnO2+2C→C)\;Sn{O_2} + 2C\xrightarrow{{}}
D)  Fe2O3+3C→250∘−400∘CD)\;Fe2{O_3} + 3C\xrightarrow{{250^\circ - 400^\circ C}}

Explanation

Solution

For solving this type of question, we need to check all the above options present, whether they give CO2CO_2 as the product in their reaction or not . By following this step, we would get our final result.So,
3C+4HNO3→Δ3CO2+4NO3+2H2O3C + 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+4HNO3→Δ  ?3C + 4HN{O_3}\xrightarrow{\Delta }\;?
So, 3C+4HNO3→ΔCO2+NO2+H2O3C + 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+4HNO3→Δ3CO2+4NO2+2H2O3C + 4HN{O_3}\xrightarrow{\Delta }3C{O_2} + 4N{O_2} + 2{H_2}O
6NaOH+2C→  ?6NaOH + 2C\xrightarrow{{}}\;?
So, 6NaOH+2C→H2+Na+Na2CO36NaOH + 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→3H2+2Na+2Na2CO36NaOH + 2C\xrightarrow{{}}3{H_2} + 2Na + 2N{a_2}C{O_3}
SnO2+2C→  ?Sn{O_2} + 2C\xrightarrow{{}}\;?
So, SnO2+2C→Sn+2COSn{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 :-
SnO2+2C→Sn+2COSn{O_2} + 2C\xrightarrow{{}}Sn + 2CO
Fe2O3+3C→250∘−400∘CFe2{O_3} + 3C\xrightarrow{{250^\circ - 400^\circ C}}
So, Fe2O3+3C→250∘−400∘CFe+COFe2{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;
Fe2O3+3C→2Fe+3COF{e_2}{O_3} + 3C\xrightarrow{{}}2Fe + 3CO
So, option (a) i.e., chemical reaction:
3C+4HNO3→Δ3CO2+4NO2+2H2O3C + 4HN{O_3}\xrightarrow{\Delta }3C{O_2} + 4N{O_2} + 2{H_2}O is the reaction which gives out CO2C{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+4HNO3→Δ3CO2+4NO2+2H2O3C + 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.