Question

Calculate the mass of water produced when 440 g of propane burns completely in air.

Answer 1

The mass of water produced when 44.0 g of propane burns completely in the air is 72.0g.

The molar masses of propane and water are $44 \mathrm{~g} / \mathrm{mol}$and $18 \mathrm{~g} / \mathrm{mol}$ respectively.

44.0 grams of propane corresponds to 1 mole of propane.

According to the balance chemical equation, $\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+$ $4 \mathrm{H}_{2} \mathrm{O}(1)$

1 mole of propane gives 4 moles of water.

Thus, the combustion of$44.0 \mathrm{~g}$ ( 1 mole) of propane will give 4 moles of water.

Mass of water obtained $=4 \times 18=72.0 \mathrm{~g}$.

The molar mass of a chemical compound is determined by dividing its mass by the quantity of that compound, expressed as the number of moles in the sample, measured in moles. The molar mass of a material is a bulk attribute rather than a molecular one.

Multiply each element's atomic weight (found in the periodic table) by the quantity of that element's atoms in the compound. 3. Add up the totals, and then follow the number with the units of grams/mole.

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