Question

2 C4H10 + 13 O2 → 8CO2 + 10 H2O Which one is the Limiting Reactant

Answer 1

Answer:

It's a lot of calculation and it took me a bit of time

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EXPLANATIONS

Consider the following combustion reaction: 2 C4H10 + 13 O2 ----> 8 CO2 + 10 H2O. 125 g or C4H10 react with 415 g of O2.

a.) Which mass of CO2 and H2O can be produced?

b.) Which substance is the limiting

a)

Molar mass of C4H10,

MM = 4*MM(C) + 10*MM(H)

= 4*12.01 + 10*1.008

= 58.12 g/mol

mass(C4H10)= 125.0 g

use:

number of mol of C4H10,

n = mass of C4H10/molar mass of C4H10

=(1.25*10^2 g)/(58.12 g/mol)

= 2.151 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 415.0 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(4.15*10^2 g)/(32 g/mol)

= 12.97 mol

Balanced chemical equation is:

2 C4H10 + 13 O2 ---> 8 CO2 + 10 H2O

2 mol of C4H10 reacts with 13 mol of O2

for 2.151 mol of C4H10, 13.98 mol of O2 is required

But we have 12.97 mol of O2

so, O2 is limiting reagent

we will use O2 in further calculation

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

According to balanced equation

mol of CO2 formed = (8/13)* moles of O2

= (8/13)*12.97

= 7.981 mol

use:

mass of CO2 = number of mol * molar mass

= 7.981*44.01

= 3.512*10^2 g

According to balanced equation

mol of H2O formed = (10/13)* moles of O2

= (10/13)*12.97

= 9.976 mol

use:

mass of H2O = number of mol * molar mass

= 9.976*18.02

= 1.798*10^2 g

Answer:

mass of CO2 = 3.51*10^2 g

mass of H20 = 1.80*10^2 g

b)

O2 is limiting reagent