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Here is the problem: In the reaction CaCO3 → CaO plus CO2, how many grams of CO2 form from 50 g of CaCO3?

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Balance atoms of every element, then use the coefficients as mole ratios.

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Write an atom balance for each element. Give each species an unknown coefficient; atoms of every element must be equal on both sides.

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Solve the linear system with the smallest whole numbers. Exact elimination leaves one free coefficient; scaling by the least common denominator gives integers.

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Balanced equation.

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Moles of CaCO3.

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CO2 via the mole ratio 1: 1.

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So the answer is CaCO3 → CaO plus CO2; 50 g CaCO3 → CO2: 21.9856 g (0.49957 mol).

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Coefficients in a balanced equation count particles (moles), not grams. Convert grams to moles, apply the mole ratio, then convert back to grams with the other species prime molar mass. Assumes the reaction goes to completion with the known species fully consumed (or formed).

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Ca: 1 equals 1, C: 1 equals 1, O: 3 equals 3, which checks out.

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CO2 is produced in a 1: 1 mole ratio to CaCO3.