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1. Naci 2. KMno, 3. Mg(NO.) 4. Calculate the percent composition of carbon in Caco,. 5. How many grams of oxygen can be produced from the decomposition of 100.0 g of KCIO,? 6. How much iron can be recovered from 25.0g of Fe,0,?

1. Naci 2. KMno, 3. Mg(NO.) 4. Calculate the percent composition of carbon in Caco,. 5. How many grams of oxygen can be produced from the decomposition of 100.0 g of KCIO,? 6. How much iron can be recovered from 25.0g of Fe,0,?

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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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10:51 * :A 0 40% Ch26 Percent Comp... Name Date Ch 26 - Percent Composition Practice Determine the percentage composition of each of the compounds below. 1. Naci 2. KMno, 3. Mg(NO:) 4. Calculate the percent composition of carbon in CaCo,. 5. How many grams of oxygen can be produced from the decomposition of 100.0 g of KCIOo,? 6. How much iron can be recovered from 25.0g of Fe;0,? 7. A chemistry student took 10.0g of an unknown hydrate and heated it in order to drive off the water of hydration. The mass of the anhydrous salt (what's left after the water is gone) was 5.83 g. What was the percent water in the original hydrate sample?
Transcribed Image Text:10:51 * :A 0 40% Ch26 Percent Comp... Name Date Ch 26 - Percent Composition Practice Determine the percentage composition of each of the compounds below. 1. Naci 2. KMno, 3. Mg(NO:) 4. Calculate the percent composition of carbon in CaCo,. 5. How many grams of oxygen can be produced from the decomposition of 100.0 g of KCIOo,? 6. How much iron can be recovered from 25.0g of Fe;0,? 7. A chemistry student took 10.0g of an unknown hydrate and heated it in order to drive off the water of hydration. The mass of the anhydrous salt (what's left after the water is gone) was 5.83 g. What was the percent water in the original hydrate sample?
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Q6)

Iron to be recovered from 25.0 g Fe2O3

                                 Molar mass of Fe2O3 = (2*At. wt. of Fe + 3 * At. wt of O)

                                                                   = 2(55.85) + 3(16)

                                                                   = 111.7 + 48

                                                                   = 159.7 g/mol

            Iron present in one mole of Fe2O3 = 111.7 g

  Iron to be recovered from 159.7 g Fe2O3 = 111.7 g

    Iron to be recovered from 25.0 g Fe2O3 => (25.0 g * 111.7 g)/159.7 g

                                                                    = 17.4859 g

Iron to be recovered from 25.0 g Fe2O3 = 17.5 g

 

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