Map Sapling Learning macmillan learning A sample of 9.45 g of liquid 1-propanol,
ID: 589624 • Letter: M
Question
Map Sapling Learning macmillan learning A sample of 9.45 g of liquid 1-propanol, C3HgO, is combusted with 24.9 g of oxygen gas. Carbon dioxide and water are the products Enter the balanced chemical equation for the reaction. Physical states are optional and not graded. Tip: If you need to clear your work and reset the equation, click the button that looks like two arrows. What is the limiting reactant? O 1-propanol O oxygen How many grams of CO2 are released in the reaction? Number How many grams of the excess reactant remain after the reaction is complete? NumberExplanation / Answer
1)
Balanced chemical equation is:
2 C3H8O + 9 O2 --> 6 CO2 + 8 H2O
2)
Molar mass of C3H8O,
MM = 3*MM(C) + 8*MM(H) + 1*MM(O)
= 3*12.01 + 8*1.008 + 1*16.0
= 60.094 g/mol
mass(C3H8O)= 9.45 g
use:
number of mol of C3H8O,
n = mass of C3H8O/molar mass of C3H8O
=(9.45 g)/(60.094 g/mol)
= 0.1573 mol
Molar mass of O2 = 32 g/mol
mass(O2)= 24.9 g
use:
number of mol of O2,
n = mass of O2/molar mass of O2
=(24.9 g)/(32 g/mol)
= 0.7781 mol
Balanced chemical equation is:
2 C3H8O + 9 O2 ---> 6 CO2 + 8 H2O
2 mol of C3H8O reacts with 9 mol of O2
for 0.1573 mol of C3H8O, 0.7076 mol of O2 is required
But we have 0.7781 mol of O2
so, C3H8O is limiting reagent
Answer: 1-propanol
3)
we will use C3H8O 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 = (6/2)* moles of C3H8O
= (6/2)*0.1573
= 0.4718 mol
use:
mass of CO2 = number of mol * molar mass
= 0.4718*44.01
= 20.8 g
Answer: 20.8 g
4)
According to balanced equation
mol of O2 reacted = (9/2)* moles of C3H8O
= (9/2)*0.1573
= 0.7076 mol
mol of O2 remaining = mol initially present - mol reacted
mol of O2 remaining = 0.7781 - 0.7076
mol of O2 remaining = 7.048*10^-2 mol
Molar mass of O2 = 32 g/mol
use:
mass of O2,
m = number of mol * molar mass
= 7.048*10^-2 mol * 32 g/mol
= 2.26 g
Answer: 2.26 g
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