One ole of aspartame (C_14 H_18 N_2 O_5) reacts with two moles of water to produ

Question

One ole of aspartame (C_14 H_18 N_2 O_5) reacts with two moles of water to produce one mole of aspartic acid (C_4H_7NO_4), one mole of methanol (CH_3OH) and one mole of mystery molecule. a. What is the molecular formula of the mystery molecule? b. What mass of the mystery molecule is produced from 378 g of aspartame? [212 g] KO_2 is used in a closed-system breathing apparatus. It removes carbon dioxide and water from exhaled air. The reaction for the removal of water is: KO_2 + H_2O rightarrow O_2 + KOH. The KOH produced is used to remove carbon dioxide by the following reaction: KOH + CO_2 rightarrow K H CO_3. a. What mass of KO_2 produces 235 g of O_2? [696 g KO_2] b. What mass of CO_2 can be removed by 123 g of KO_2? (76.1 g CO_2] In the combustion of propane (C_3H_8), if 200.0 g of propane is burned in air, how many grams of water are produced? [326.8 g H_2O] 1.000 g of an organic molecule that contains only carbon and hydrogen burns completely in the presence of excess oxygen. It yields 0.0333 mol of CO_2 and 0.599 g of H_2O. (a) Find the empirical mass of the compound. (b) The molecular mass was determined experimentally to be 84.159 g/mol. What is the molecular formula? [C_6H_12] Balance these equations, and identify the type of reaction H_2O(l) rightarrow H2 (g) + O_2 (g) Zn (s) + Au (aq)^+ rightarrow Zn (aq)^2+ + Ag (s)

Explanation / Answer

Queestion 2a

Balanced equation:
C14H18N2O5 + 2 H2O ===> C4H7NO4 + CH4O + C9H11O2N

Molecular formula of mystery molecule is C9H11O2N and moleuclar weight is 165.19

Question 2 b

378 gm of Aspartame = 378/294.30 = 1.284 Moles

1.284 Moles of Aspartame will produce the same moles of Mystery molecule

Mass of Mystery molecule to be produced from 378 gm Aspartame = 1.284 x 165.19 = 212.16 gm

Quesiton 3a

235 gm of O2 = 235 /32 = 7.344 Moles

To produce 7.344 moles it needs ( 7.344 x 4/3) 9.7920 Moles of KO2

Mass of KO2 needed = 9.79 x 71.097 = 696.18 gm

Question 3b

123 gm of KO2 = 123 /71.097 = 1.730 Moles

1.73 Moles of KO2 will produce the same moles of KOH. Hence 1.73 Moles of KOH will react with the same moles of CO2

Moles of CO2 is needed = 1.73 Moles

Mass of CO2 = 1.73 x 44 = 76.12 gm

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