Question

Thankyou!

Consider the following balanced equation: 502(g) + C3H8(9) 3CO2(g) + 4H20(1) If 19.9 moles of O2(g) and 4.42 moles of CzHg(g) are allowed to react, what is the theoretical yield of CO2(g) in moles? 0 O OOOO 41.0 moles 94.3 moles 36.3 moles 35.3 moles 11.9 moles

Answer 1

1)
Balanced chemical equation is:
5 O2 + C3H8 ---> 3 CO2 + 4 H2O


5 mol of O2 reacts with 1 mol of C3H8
for 19.9 mol of O2, 3.98 mol of C3H8 is required
But we have 4.42 mol of C3H8

so, O2 is limiting reagent
we will use O2 in further calculation

According to balanced equation
mol of CO2 formed = (3/5)* moles of O2
= (3/5)*19.9
= 11.9 mol
Answer: 11.9 mol

2)

Molar mass of HCl,
MM = 1*MM(H) + 1*MM(Cl)
= 1*1.008 + 1*35.45
= 36.458 g/mol


mass(HCl)= 54.1 g

use:
number of mol of HCl,
n = mass of HCl/molar mass of HCl
=(54.1 g)/(36.46 g/mol)
= 1.484 mol

Molar mass of Al = 26.98 g/mol


mass(Al)= 85.8 g

use:
number of mol of Al,
n = mass of Al/molar mass of Al
=(85.8 g)/(26.98 g/mol)
= 3.18 mol
Balanced chemical equation is:
6 HCl + 2 Al ---> 3 H2 + 2 AlCl3


6 mol of HCl reacts with 2 mol of Al
for 1.484 mol of HCl, 0.4946 mol of Al is required
But we have 3.18 mol of Al

so, HCl is limiting reagent
we will use HCl in further calculation


According to balanced equation
mol of Al reacted = (2/6)* moles of HCl
= (2/6)*1.484
= 0.4946 mol
mol of Al remaining = mol initially present - mol reacted
mol of Al remaining = 3.18 - 0.4946
mol of Al remaining = 2.686 mol
Answer: 2.69 mol

Only 1 question at a time please. I have answered 1st two for you