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The solving of mole to mole problems through stoichiometry is based in ratios. Assuming the reaction formulas are simplified and balanced the moles consumed and produced can be calculated. For example in the simple combustion reaction of ethane and oxygen of:CH4 + 2O2 = CO2 +2H20 we see a molar ratio of 1:2 between methane combusted and waters produced.

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9y ago
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8y ago

1 mol is equal to the molecular mass of a compound/litre of water.

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Q: How do you solve for moles in chemistry?
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