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Mass Relationships in Chemical Reactions Chapter 3
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Empirical Formula of a Hydrate Hydrated copper (II) sulfate has the formula CuSO 4 · x H 2 O. To find x, 1.023 g of the blue solid is heated in a crucible until its mass no longer decreases. The mass of the anhydrous, white CuSO 4 is 0.654 g. How many moles of water ( x ) are there per mole of CuSO 4 ?
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Mass Changes in Chemical Reactions a.k.a. “Stoichiometry”
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Methanol burns in air according to the equation 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH 3 OHmoles CH 3 OHmoles H 2 Ograms H 2 O molar mass CH 3 OH coefficients chemical equation molar mass H 2 O 209 g CH 3 OH 1 mol CH 3 OH 32.0 g CH 3 OH x 4 mol H 2 O 2 mol CH 3 OH x 18.0 g H 2 O 1 mol H 2 O x = 235 g H 2 O
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Given the following balanced equation, 2Mg + O 2 → 2MgO (a) how many grams of oxygen are required to completely react with 0.145 g of magnesium? (b) How many grams of magnesium oxide are formed?
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Limiting Reagent Reactant that is present in the smaller or smallest required stoichiometric ratio Fig 3.15 2 H 2 (g) + O 2 (g) → 2 H 2 O (v)
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Fig 3.15 2 H 2 (g) + O 2 (g) → 2 H 2 O (v) i.e., in this case O 2 is in excess and H 2 is the limiting reagent (LR) The amount of product depends on LR alone!!
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How do we determine the limiting reagent? Calculate the theoretical yield twice: Assume first reagent is LR and calculate Assume second reagent is LR and calculate The smaller result gives LR and yield
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How many grams of water will be formed from 150 g H 2 and 1500 g O 2 in a fuel cell? Sample exercise 3.19 p 104 2 H 2 (g) + O 2 (g) → 2 H 2 O (v) Ans: 1400 g H 2 O This is the theoretical yield
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Theoretical yield - the maximum amount of product that can be made –In other words it’s the amount of product possible as calculated through the stoichiometry problem. This is different from the actual yield, which is the amount one actually produces and measures. Actual Yield Theoretical Yield Percent Yield =x 100%
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How many grams of water will be formed from 150 g H 2 and 1500 g O 2 in a fuel cell? Sample exercise 3.19 p 104 2 H 2 (g) + O 2 (g) → 2 H 2 O (v) Ans: 1400 g H 2 O This is the theoretical yield Assume, say, 1250 g H 2 O are formed. Then percent yield =
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