Bishop_Book_10_eBook - C HAPTER 10 C HEMICAL C ALCULATIONS...

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Unformatted text preview: C HAPTER 10 C HEMICAL C ALCULATIONS AND C HEMICAL E QUATIONS 367 lthough Chapter 9 was full of questions that began with, “ How much…? ” we are not done with such questions yet. In Chapter 9, our questions focused on chemical formulas. For example, we answered such questions as, “ How much of the element vanadium can be obtained from 2.3 metric tons of the compound V 2 O 5 ?” The chemical formula V 2 O 5 told us that there are two moles of vanadium, V, in each mole of V 2 O 5 . We used this molar ratio to convert from moles of V 2 O 5 to moles of vanadium. In this chapter, we encounter questions that focus instead on chemical reactions. These questions ask us to convert from amount of one substance in a given chemical reaction to amount of another substance participating in the same reaction. For example, a business manager, budgeting for the production of silicon‑based computer chips, wants to know how much silicon can be produced from 16 kg of carbon and 32 kg of silica, SiO 2 , in the reaction SiO 2 ( s ) + 2C( s ) Si( l ) + 2CO( g ) 2000 ° C A safety engineer in a uranium processing plant, wants to know how much water needs to be added to 25 pounds of uranium hexafluoride to maximize the synthesis of UO 2 F 2 by the reaction UF 6 + 2H 2 O → UO 2 F 2 + 4HF A chemistry student working in the lab might be asked to calculate how much 1‑bromo‑2‑methylpropane, C 4 H 9 Br, could be made from 6.034 g of 2‑methyl‑2‑propanol, C 4 H 9 OH, in the reaction 3C 4 H 9 OH + PBr 3 → 3C 4 H 9 Br + H 3 PO 3 In these calculations, we will be generating conversion factors from the coefficients in the balanced chemical equation. 10.1 Equation Stoichiometry 10.2 Real-World Applications of Equation Stoichiometry 10.3 Molarity and Equation Stoichiometry How much product can be made from the given reactants? 10.1 Equation Stoichiometry Chapter 9 asked you to pretend to be an industrial chemist at a company that makes phosphoric acid, H 3 PO 4 . The three‑step “furnace method” for producing this compound is summarized by these three equations: 2Ca 3 (PO 4 ) 2 + 6SiO 2 + 10C → 4P + 10CO + 6CaSiO 3 4P( s ) + 5O 2 ( g ) → P 4 O 10 ( s ) P 4 O 10 ( s ) + 6H 2 O( l ) → 4H 3 PO 4 ( aq ) Following the strategy demonstrated in Example 9.6, we calculated the maximum mass of tetraphosphorus decoxide, P 4 O 10 , that can be made from 1.09 × 10 4 kilograms of phosphorus in the second of these three reactions. The answer is 2.50 × 10 4 kg P 4 O 10 . We used the following steps: Balance chemical equations. (Section 4.1.) Write or identify the definitions of solution , solute , and solvent . (Section 4.2) Given a description of a solution, identify the solute and solvent. (Section 4.2) Describe water solutions in terms of the nature of the particles in solution and the attractions between them. (Sections 4.2 and 5.1) Given formulas for two ionic compounds, predict whether a precipitate will form when water solutions of the two are mixed, and write the...
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This note was uploaded on 03/03/2012 for the course CHEM 100 taught by Professor Mark during the Fall '06 term at Monterey Peninsula College.

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Bishop_Book_10_eBook - C HAPTER 10 C HEMICAL C ALCULATIONS...

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