Some Basic Concepts NEET 2026: Butane CO₂ Calculation

What is the answer to the Some Basic Concepts NEET 2026 n-butane question?

The correct answer is A) 352 g of carbon dioxide. This Some Basic Concepts NEET 2026 question asks you to calculate the mass of CO₂ obtained when 116 g of n-butane burns completely. The source is NEET 2026, Code-50, Chemistry, Some Basic Concepts in Chemistry.

The supplied atomic masses, in amu, are:

H = 1, C = 12, O = 16

The answer choices are:

“Medium” difficulty and “75 seconds” are question-bank tags, not official exam classifications or a guaranteed solving time. The required conversion chain is fuel mass, fuel moles, product moles, then product mass.

How do you balance n-butane combustion and find the mole ratio?

One mole of n-butane produces four moles of CO₂ on complete combustion. The balanced equation gives this mole ratio, not a mass ratio. Treat oxygen as sufficient because the question specifies complete combustion and gives no oxygen-limited calculation.

The molecular formula of n-butane is: $C_{4} H_{10}$

Complete combustion produces carbon dioxide and water. Balance carbon and hydrogen first; this intermediate equation is not yet oxygen-balanced:

C_{4} H_{10} + O_{2} \to 4 C O_{2} + 5 H_{2} O

Count the oxygen atoms on the product side:

4 \times 2 + 5 \times 1 = 13 oxygen atoms

Each oxygen molecule contains two oxygen atoms, so its coefficient must be: $\frac{13}{2}$

C_{4} H_{10} + \frac{13}{2} O_{2} \to 4 C O_{2} + 5 H_{2} O

Multiply every coefficient by two to obtain the official equation:

2 C_{4} H_{10} + 13 O_{2} \to 8 C O_{2} + 10 H_{2} O

Read the mole relationship:

2 m o l C_{4} H_{10} \to 8 m o l C O_{2}

1 m o l C_{4} H_{10} \to 4 m o l C O_{2}

Coefficients compare numbers of molecules or moles, not masses in grams. Converting between masses also requires molar masses.

How does 116 g of n-butane become 8 mol of CO₂?

The sample contains two moles of butane, which produce eight moles of CO₂. Keep the sample-mole calculation separate from the product-mole calculation. Jumping from the equation’s one-mole relationship straight to product mass loses the factor of two.

First calculate butane’s molar mass. Its carbon contribution is:

4 \times 12 = 48 g m o l^{- 1}

Its hydrogen contribution is:

10 \times 1 = 10 g m o l^{- 1}

Add them:

M (C_{4} H_{10}) = 48 + 10 = 58 g m o l^{- 1}

Convert the supplied sample into moles: $n = \frac{m}{M}$

n (C_{4} H_{10}) = \frac{116 g}{58 g m o l^{- 1}} = 2 m o l

Apply the balanced-equation ratio:

n (C O_{2}) = 2 m o l C_{4} H_{10} \times \frac{4 m o l C O_{2}}{1 m o l C_{4} H_{10}} = 8 m o l C O_{2}

Four is the number of moles of CO₂ per mole of butane. It is not the total CO₂ amount for every sample size.

How do you convert 8 mol of CO₂ into grams and check answer A?

Eight moles of CO₂ have a mass of 352 g, confirming A. The product mass exceeds the starting fuel mass because oxygen from the surroundings becomes part of the products. Conservation of mass applies to all reactants and products, not to fuel and CO₂ alone.

Calculate the CO₂ molar mass:

M (C O_{2}) = 12 + 2 \times 16 = 44 g m o l^{- 1}

Then convert moles into mass: $m = n M$

m (C O_{2}) = 8 m o l \times 44 g m o l^{- 1} = 352 g

Final answer: A) 352 g. The full conversion chain below shows how each unit cancels; read the lines as one continuous product:

& 116 g C_{4} H_{10} \times \frac{1 m o l C_{4} H_{10}}{58 g C_{4} H_{10}} & \times \frac{4 m o l C O_{2}}{1 m o l C_{4} H_{10}} \times \frac{44 g C O_{2}}{1 m o l C O_{2}} & = 352 g C O_{2}

Use carbon conservation as an independent check, after the mole-based solution. All the carbon in the burned fuel must appear in the CO₂.

The fuel contains:

m_{carbon in fuel} = 2 \times 4 \times 12 = 96 g

The calculated CO₂ contains:

m_{{carbon in CO}_{2}} = 352 \times \frac{12}{44} = 96 g

Both carbon masses agree. This checks the answer without replacing the official method.

Why does option C give 176 g instead of 352 g?

176 g is correct for 58 g of butane, not for the stated 116 g. One way to obtain option C is to read the one-mole relationship correctly but never scale it to the actual sample. This is a sample-size mistake, not a balancing mistake.

The incomplete route is:

1 m o l b u t a n e \to 4 m o l C O_{2}

4 m o l \times 44 g m o l^{- 1} = 176 g

The omitted conversion is:

\frac{116 g}{58 g m o l^{- 1}} = 2 m o l b u t a n e

Compare the routes directly; each starts with a different fuel mass:

58 g b u t a n e & \to 1 m o l b u t a n e & \to 4 m o l C O_{2} \to 176 g C O_{2}

116 g b u t a n e & \to 2 m o l b u t a n e & \to 8 m o l C O_{2} \to 352 g C O_{2}

Repair rule: write the supplied reactant amount in moles before applying any coefficient ratio. The equation gives a proportion; the sample mass determines how much product forms.

Can you apply the same method to three related questions?

Use the same balanced equation and supplied atomic masses for all three variations below. These are newly written practice questions, not additional NEET PYQs. They test changing the sample size, choosing a different coefficient ratio and reversing the conversion chain.

Attempt all three prompts before reading the worked answers:

What mass of CO₂ forms from complete combustion of 58 g of n-butane?
What mass of O₂ is consumed when 116 g of n-butane burns completely?
What mass of n-butane must burn completely to produce 88 g of CO₂?

Answer 1: changing the sample size. The coefficient ratio stays the same; only the starting amount changes.

n (C_{4} H_{10}) = \frac{58 g}{58 g m o l^{- 1}} = 1 m o l

n (C O_{2}) = 1 m o l C_{4} H_{10} \times \frac{4 m o l C O_{2}}{1 m o l C_{4} H_{10}} = 4 m o l C O_{2}

m (C O_{2}) = 4 m o l \times 44 g m o l^{- 1} = 176 g

Answer 2: selecting a different coefficient ratio. Use the butane-to-oxygen ratio, not the butane-to-CO₂ ratio.

n (C_{4} H_{10}) = \frac{116 g}{58 g m o l^{- 1}} = 2 m o l

n (O_{2}) = 2 m o l C_{4} H_{10} \times \frac{13 m o l O_{2}}{2 m o l C_{4} H_{10}} = 13 m o l O_{2}

M (O_{2}) = 2 \times 16 = 32 g m o l^{- 1}

m (O_{2}) = 13 m o l \times 32 g m o l^{- 1} = 416 g

Answer 3: reversing the conversion chain. Start with product moles, then divide by four to find fuel moles.

n (C O_{2}) = \frac{88 g}{44 g m o l^{- 1}} = 2 m o l

n (C_{4} H_{10}) = 2 m o l C O_{2} \times \frac{1 m o l C_{4} H_{10}}{4 m o l C O_{2}} = 0.5 m o l C_{4} H_{10}

m (C_{4} H_{10}) = 0.5 m o l \times 58 g m o l^{- 1} = 29 g

Now cover the answers and redo Question 3. Write the species beside each mole quantity so the direction of the ratio stays visible.

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Frequently asked questions

How much CO₂ is produced by complete combustion of 116 g of n-butane?

Complete combustion of 116 g of n-butane produces 352 g of CO₂, so the correct option is A. Butane has a molar mass of 58 g/mol, making the sample 2 mol. Each mole of butane produces 4 mol of CO₂, giving 8 × 44 = 352 g.

What is the balanced equation for complete combustion of n-butane?

The balanced equation is 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O. This gives a butane-to-CO₂ mole ratio of 1:4. The coefficients represent mole ratios, not mass ratios.

Why is 176 g the wrong answer for the 116 g butane question?

176 g of CO₂ corresponds to burning 58 g, or 1 mol, of butane completely. The supplied 116 g sample contains 2 mol of butane and therefore produces twice that mass of CO₂: 352 g. Convert the supplied fuel mass into moles before applying the equation's mole ratio.

Why does the CO₂ produced weigh more than the butane burned?

Oxygen consumed during combustion contributes to the mass of the CO₂ produced. Conservation of mass applies to all reactants and products, including oxygen and water, not to butane and CO₂ alone.

Some Basic Concepts NEET 2026: n-Butane Combustion