Question

222 g of MTBE (CO(CH3)4) are added to gasoline, resulting in a total volume of 2 L of reformulated gas (RFG). Assume that the density of RFG is 0.70 g/mL, and the density of MTBE is 0.74 g/mL. For a-d, determine the concentration of MTBE in the RFG in the following units?
a. mg/L
b. moles/L
c. % (w/w)
d. % (v/v)
e. What is the % (w/w) oxygen in the RFG due to MTBE?

Answer 1

Assume That The Density Of RFG Is 0.7 G/mL And The Density Of MTBE Is ... are added to gasoline, resulting in a total volume of 2 L of reformulated gas (RFG). ... Assume that the density of RFG is 0.7 g/mL and the density of MTBE is 0.74 g/mL. Determine the concentration of MTBE in the RFG in the following units (a-d)

Answer 2

Answer:

Explanation:

From the information given:

(a)

$Concentration \ in \ mg/L = \dfrac{Mass \ of \ MTBE \ in \ mg}{Total \ volume (in \ L)}$

$Concentration \ in \ mg/L = \dfrac{222 \times 10^3 \ mg}{22}$

$Concentration \ in \ mg/L = 111 \times 10^3 \ mg/L$

(b)

$number \ of \ mole s= \dfrac{mass}{molar \ mass } \\ \\ number \ of \ mole s=\dfrac{222 \ g}{88.15 \ g/mol} \\ \\ \mathbf{= 2.518 mol}$

(c)

$w/w \ percentage = \dfrac{mass \ of \ MTBE }{mass \ of \ solution (RFG)}\times 100\%$

$where; \\ \\ mass \ of \ (RFG) = 2L \times 0.70 g/mL \\ \\ mass \ of \ (RFG) = 2000 ml \times 0.70 g/mL \\ \\ mass \ of \ (RFG) = 1400 g$

∴

$w/w \ percentage = \dfrac{222 \ g}{1400 \ g}\times 100\% = \mathbf{15.8\%}$

(d)

$Volume of MTBE =\dfrac{mass \ of \ MTBE}{density \ of \ MTBE}$

$Volume \ of \ MTBE = 300 \ mL$

∴

$v/v\% = \dfrac{volume \ of \ MTBE}{volume \ of \ RFG} \\ \\ v/v\% =\dfrac{300 \ mL}{2000 \ mL}\times 100\% \\ \\ \mathbf{v/v\% = 15.00\%}$

(e)

$From \the \ given \ information; \\ \\ 2.5184 \ moles\ of \ MTBE contain \ 2.5184 \ mole of oxygen$

∴

$mass of oxygen MTBE = 2.5284 mol \times 16\ g/mol \\ \\ mass of oxygen MTBE = 40.3 9 \ g\\ \\ mass\ of \ RFG = 1400 g$

∴

$\% w/w = \dfrac{mass \ of \ oxygen}{mass \ of RFG }=\dfrac{40.22 \ g}{1400 \ g} \times 100\%$

$\% w/w == 2.88\%$