1. For the following reaction, 5.70 grams of propane (C3H8) are allowed to react
ID: 1036512 • Letter: 1
Question
1. For the following reaction, 5.70 grams of propane (C3H8) are allowed to react with 24.2 grams of oxygen gas.
2.
For the following reaction, 0.154 moles of barium hydroxide are mixed with 0.198 moles ofsulfuric acid.
For the following reaction, 5.70 grams of propane (CzHg) are allowed to react with 24.2 grams of oxygen gas. propane (CzHg) (g) + oxygen (g) »carbon dioxide (g) + water (g) What is the maximum amount of carbon dioxide that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams Submit Answer Try Another Version 2 item attempts remainingExplanation / Answer
1)
a)
Molar mass of C3H8,
MM = 3*MM(C) + 8*MM(H)
= 3*12.01 + 8*1.008
= 44.094 g/mol
mass(C3H8)= 5.7 g
use:
number of mol of C3H8,
n = mass of C3H8/molar mass of C3H8
=(5.7 g)/(44.094 g/mol)
= 0.1293 mol
Molar mass of O2 = 32 g/mol
mass(O2)= 24.2 g
use:
number of mol of O2,
n = mass of O2/molar mass of O2
=(24.2 g)/(32 g/mol)
= 0.7562 mol
Balanced chemical equation is:
C3H8 + 5 O2 ---> 3 CO2 + 4 H2O
1 mol of C3H8 reacts with 5 mol of O2
for 0.1293 mol of C3H8, 0.6463 mol of O2 is required
But we have 0.7562 mol of O2
so, C3H8 is limiting reagent
we will use C3H8 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 = (3/1)* moles of C3H8
= (3/1)*0.1293
= 0.3878 mol
use:
mass of CO2 = number of mol * molar mass
= 0.3878*44.01
= 17.07 g
Answer: 17.1 g
b)
C3H8
c)
According to balanced equation
mol of O2 reacted = (5/1)* moles of C3H8
= (5/1)*0.1293
= 0.6463 mol
mol of O2 remaining = mol initially present - mol reacted
mol of O2 remaining = 0.7562 - 0.6463
mol of O2 remaining = 0.1099 mol
Molar mass of O2 = 32 g/mol
use:
mass of O2,
m = number of mol * molar mass
= 0.1099 mol * 32 g/mol
= 3.517 g
Answer: 3.52 g
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