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Calculate the number of grams of Fe formed when 0.300kg of Fe2O3 reacts. Equation: Fe2O3+CO-->Fe+CO2 converted .300kg to 300g Balanced Fe2O3+3CO-->2Fe+3CO2 So I started by trying this: 300gFe2O3(1molFe2O3/159.687gFe2O3)...and then
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Equation: Fe2O3 + 2Al ---> 2Fe + Al2O3. How many grams of Fe2O3 react with excess Al to make 475g Fe? Can you show me step by step how to do it? My book says the answer is 679g Fe2O3.
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A sample weighing 3.064 g is a mixture of Fe2O3 (molar mass = 159.69) and Al2O3 (molar mass = 101.96). When heat and a stream of H2 gas is applied to the sample, the Fe2O3 reacts to form metallic Fe and H2O(g). The Al2O3 does not
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Please tell me what I did wrong on this problem: Given the equation Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g), Calculate the number of grams of CO that can react with 0.370kg of Fe2O3 Here's what I did: .370kg x 1000 = 370g of Fe2O3
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Fe2O3(s) + 2Al(s)-> 2Fe(l) + Al2O3(s) What is the theoretical yield of iron if we begin the reaction with 10.5 g of Al (assume an excess of Fe2O3)
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