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30.8g of iron (approximately) reacted with 13.2g of oxygen will yield 44g of iron(III) oxide (Fe2O3) with 2.8g of oxygen left unreacted. This assumes Atomic Mass numbers of 56 and 16 respectively for iron and oxygen. The actual mass number of iron is 55.847 and oxygen 15.9994 making the figures 30.775g of iron and 13.225g of oxygen with 2.775g of unreacted oxygen. Of course this is an exothermic reaction so will there be a tiny tiny loss of mass in the system as it is converted to heat energy, according to E=MC^2? I'll let you work that one out...

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17y ago
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2d ago

Iron(III) oxide is composed of one iron atom and three oxygen atoms. Therefore, if we have 44 grams of iron(III) oxide, we can calculate the mass of iron by subtracting the mass of oxygen. 44g - 16g (mass of oxygen) = 28g of iron is needed to combine with 16g of oxygen to form 44g of iron(III) oxide.

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Q: How many grams of iron must be added to 16 grams of oxygen to equal 44 grams of iron 3 oxide?
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