Ammonia NH 3 reacts with the oxygen in the air to produce nitrogen monoxide and

Ammonia nh 3 reacts with the oxygen in the air to

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Ammonia (NH 3 ) reacts with the oxygen in the air to produce nitrogen monoxide and water. Write balanced equations for each of these reactions. 60
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Limiting Reactants Limiting reactant the reactant that is consumed first and therefore limits the amounts of products that can be formed. Determine which reactant is limiting to calculate correctly the amounts of products that will be formed. 61
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Limiting Reactants Methane and water will react to form products according to the equation: CH 4 + H 2 O 3H 2 + CO 62 CH 4 + H 2 O react in a 1: 1 ratio If I react 1 mole of CH 4 with 1 mole of H 2 O, how many moles of CO do I make? If I react 2 moles of CH 4 with 1 mole of H 2 O, how many moles of CO do I make?
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Mixture of 6CH 4 and 9H 2 O Molecules Reacting 63 Reaction stops once the methane is all gone limiting reagent
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Limiting Reactants: reacting 6CH 4 with 9H 2 O The amount of products that can form is limited by the methane. Methane is the limiting reactant. Water is in excess. 64
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Note We cannot simply add the total moles of all the reactants to decide which reactant mixture makes the most product. We must always think about how much product can be formed by using what we are given, and the ratio in the balanced equation. 65
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66 LIMITING REAGENT FLOW CHART 1. Write the reaction 2. Balance the reaction 3. Convert all masses to grams 4. Convert grams to moles 5. Determine the limiting reagent 6. Find the number of moles of product from the limiting reagent 7. Convert moles of product to grams of product 8. Did you answer the right question? 9. Does your answer make sense? moles moles grams grams ---------------------------------------------------------------------------------------- REACTANT SIDE PRODUCT SIDE
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67 Consider the reaction phosphorous (P 4 ) with fluorine to produce phosphorus trifluoride. How much phosphorous trifluoride can be produced by 85.0 g of P 4 reacting with 191g F 2 ? 1. Write the reaction: P 4 + F 2 PF 3 2. Balance: P 4 + 6F 2 4PF 3 3. Convert all masses to grams (done) P is 31amu F is 19amu
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68 Consider the reaction phosphorous (P 4 ) with fluorine to produce phosphorus trifluoride. How much phosphorous trifluoride can be produced by 85.0 g of P 4 reacting with 191g F 2 ? 4.Convert grams to moles 4 685 . 0 124 1 0 . 85 molP g mol g 2 02 . 5 0 . 38 1 191 molF g mol g
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69 5. Determine the limiting reagent: P 4 + 6F 2 4PF 3 0.685mol 5.02mol Predict product:
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70 5. Determine the limiting reagent: (1) P 4 + 6F 2 4 PF 3 0.685mol 5.02mol Predict product: x = 2.74mol PF 3 685 . 0 1 4 4 3 x molP molPF
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71 5. Determine the limiting reagent: P 4 + 6 F 2 4 PF 3 0.685mol 5.02mol Predict product: x = 2.74mol PF 3 x = 3.35mol PF 3 P 4 predicts less product, so it is the limiting reagent, and 2.74mol PF 3 is the theoretical yield. 685 . 0 1 4 4 3 x molP molPF 02 . 5 6 4 2 3 x molF molPF
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72 6. Find the number of moles of product from the limiting reagent (done: 2.74mol) 7. Convert moles product to grams product g mol g mol 241 1 0 . 88 74 . 2 8. Did you answer the right question?
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