Question

I need help with Questions 14, 17, and 18. Please show work!

of 6 + ... JE SAZ 14) What mass of gaseous carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide (according to the unbalanced equation below)? LiOH(s) + CO2(g) → LiCO, (s) + H2O (1) a. 920.8 b. 1.00 kg c. 1840 g d. 2.00 kg 15) Aspartame in as artificial sweetener that is 160 times sweeter than sucrose (table sugar) when dissolved in water. It is marketed at Nutra-Sweet. The molecular formula of aspartame is C4H1N,Os. Calculate the molar mass of aspartame. a. 276.16 g b. 280.29 g c. 230.30 g d. 294.30 g 16) Adipic acid is an organic compound composed of 49.31% C, 43.79% 0, and the rest is hydrogen. If the molar mass of adipic acid is 146.1 g/mol, what is the molecular formula of adipic acid?

Answer 1

14.

Correct option: a

Mass of LiOH = 1.00 kg = 1000 g

Molar mass of LiOH = 23.95 g/mol

Moles of LiOH = 1000 g/23.95 g/mol = 41.8 mol

Balanced equation:

2LiOH + CO₂$\rightarrow$Li₂CO₃ + H₂O

Theoretically, 2 mol of LiOH absorb 1 mol of CO₂

Therefore, 41.8 mol of LiOH absorb = (1 mol/2 mol) x 41.8 mol = 20.9 mol of CO₂

Molar mass of CO₂ = 44 g/mol

Hence, the mass of CO₂ can be absorbed = 20.9 mol x 44 g/mol = 920 g

17.

Correct option: c

Mass of butane = 100 g

Molar mass of butane = 58.12 g/mol

Moles of butane = 100 g/ 58.12 g/mol = 1.72 mol

Mass of oxygen = 100 g

Molar mass of oxygen = 32 g/mol

Moles of oxygen = 100 g/ 32 g/mol = 3.12 mol

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

For complete reaction, the required mole ratio of butane : O₂ = 2 : 13

The available mole ratio of butane : O₂ = 1.72 : 3.12

                                                               = 1 : 1.8

                                                               = 2 : 3.6

Thus, per mol of butane, the available mol of O₂ is less than required.

Hence, butane is the limiting reagent in the reaction.

Theoretically, 2 mol of butane yields 8 mol of CO₂

Therefore, 1.72 mol of butane yields = (8 mol/2 mol) x 1.72 mol = 6.88 mol of CO₂

Molar mass of CO₂ = 44 g/mol

Hence, the theoretical yield of CO₂ = 6.88 mol x 44 g/mol = 303 g

Actual yield of CO₂ = 50.0 g

Percent yield = (50.0 g/303 g) x 100 = 16.5 %

18.

Correct option: c

Balanced molecular equation:

Fe(NO₃)₃(aq) + 3KOH(aq) $\rightarrow$ Fe(OH)₃(s) + 3KNO₃(aq)

The complete ionic equation:

Fe³⁺(aq) + 3NO₃^-(aq) + 3K⁺(aq) + 3OH^-(aq) $\rightarrow$ Fe(OH)₃(s) + 3K⁺(aq) + 3NO₃^-(aq)

Hence, the coefficient in front of nitrate ion = 3