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Iron combines with 5 g of Copper (1) Nitrate to form 6.01 g of Iron (1) Nitrate and 0.4 g of copper metal. How much Iron was required to complete this reaction?
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Apr 2, 2022
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Iron combines with 5 g of Copper (1) Nitrate to form 6.01 g of Iron (1) Nitrate and 0.4 g of copper
metal. How much Iron was required to complete this reaction?
Chemistry
high-school
Elfif
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First, you would add the products: 0.4 + 6.01 = 6.41 g
Then, to get the mass of Iron required, you would do 6.41 - 5.0 = 1.41
This is because in the conservation of mass principle it states,
mass of reactants = mass of products.
Joe Martinez
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Apr 9, 2022
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Joe Martinez
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