± Core Chemistry Skill: Calculating Mass of Product from a Limiting Reactant Con

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± Core Chemistry Skill: Calculating Mass of Product from a Limiting Reactant Consider a siluation in which 186 g of P4 are exposed to 208 g of O2 Part A Learning Goal: What is the maximum number of moles of P2Os that can th Express your answer to three significant figures and include the appropriate units. be made from 186 g of P4 and excess oxygen? To calculate theoretical and percent yields, given the masses of multiple reactants. A balanced chemical reaction gives the mole ratios between reactants and products. For example, one mole of white phosphorus, P. reacts with five moles of oxygen, O2, to produce two moles of diphosphorus pentoxide, P205 Hints 3.00 mol Submit My Answers Give Up In this example, if we know the number of moles of phosphorus, we know that twice that much product can be made, assuming we have enough oxygen. Similarly, if we know the number of moles of oxygen, we know that two-fifths that much product can be made, assuming we have enough phosphorus. In a situation where you are given the mole amounts of both reactants, you should do this type of calculation individually for each one, finding two different values for the amount of product. The smaller of these two values is the maximum amount of product that can be made, also known as the theoretical yield All attempts used; correct answer displayed Part B What is the maximum number of moles of P2Os that can th be made from 208 g of 02 and excess phosphorus? Express your answer to three significant figures and include the appropriate units. If you are given the masses of the reactants, you will first have to convert to moles. Once you have mole amounts, you can follow the procedure described above, applying the ratios shown in the balanced chemical equation. Hints 208 Submit My Answers Give Up Incorrect; Try Again; 4 attempts remaining

Explanation / Answer

For the reaction,

P4 + 5O2 --> 2P2O5

part A) moles P4 = 186 g/124 g/mol = 1.5 mol

moles P2O5 formed = 2 x 1.5 = 3.0 mol

Part B) moles O2 = 208 g/32 g/mol = 6.5 mol

moles P2O5 formed = 6.5 x 2/5 = 2.6 mol

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7.72) for the reaction,

C2H6O + 3O2 --> 2CO2 + 3H2O

Part A) moles C2H6O = 6.55 g/46.07 g/mol = 0.142 mol

grams O2 needed = 3 x 0.142 mol x 32 g/mol = 13.65 g

Part B) moles H2O produced = 0.142 mol x 3

grams H2O produced = 3 x 0.142 mol x 18 g/mol = 7.67 g

Part C) moles CO2 produced = 2 x 0.142 mol

grams CO2 produced = 2 x 0.142 mol x 44 g/mol = 12.50 g

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