Finds how many grams of product a given mass of reactant yields. The mass is converted to an amount in moles, scaled by the ratio of the coefficients in the balanced equation, and converted back to a mass. Masses themselves do not follow the coefficient ratio.
The coefficients in a chemical equation give the ratio in which particles react. They are not a ratio of masses, and treating them as one produces wrong answers.
That is why there is no direct route from mass to mass. The mass is converted into an amount in moles, the coefficient ratio is applied there, and the result is converted back into a mass.
The coefficient ratio applies only in the middle step, between the two amounts.
Take the combustion of methane, , and ask how much water 16 g of methane produces.
Methane has a molar mass of 16.04, so the amount is , about 0.9975 mol.
The equation shows that each methane molecule yields two water molecules, so the water is twice that amount, about 1.9950 mol.
Water has a molar mass of 18.02, giving , about 35.95 g.
16 g of methane produced 36 g of water, more than we started with.
Nothing is wrong. Oxygen from the air joined in. The reactants total 16.04 of methane plus 64.00 of oxygen, or 80.04, and the products total 44.01 of carbon dioxide plus 36.04 of water, or 80.05, agreeing to within rounding. Mass is conserved across the reaction as a whole, not substance by substance.
This calculation assumes the reactant entered is consumed entirely. In practice whichever reactant runs out first, the limiting reagent, decides how much product forms. Run the calculation for each reactant and the smallest answer is the real one.
The equation must already be balanced, with the same count of every atom on both sides. Applying coefficients from an unbalanced equation gives a wrong answer no matter how carefully the arithmetic is done.
What comes out is the theoretical maximum. Real reactions give less, and the yield tool here computes that fraction.