Question

1. When potassium chlorate is oxidized with sucrose, the products are potassium chloride, carbon dioxide, and water. You begin with 25.5 grams of potassium chlorate and 13.4 grams of sucrose.

a. What is the limiting reactant?

b. What mass of KCl is produced?

c. How much of the excess is left over after the reaction occurs?

2. In 274 grams of Ca(NO3)NH4NO3· 3H2O, how many grams of that would be nitrogen?

Answer 1

1)

Molar mass of C12H22O11,
MM = 12*MM(C) + 22*MM(H) + 11*MM(O)
= 12*12.01 + 22*1.008 + 11*16.0
= 342.296 g/mol


mass(C12H22O11)= 13.4 g

number of mol of C12H22O11,
n = mass of C12H22O11/molar mass of C12H22O11
=(13.4 g)/(342.296 g/mol)
= 3.915*10^-2 mol

Molar mass of KClO3,
MM = 1*MM(K) + 1*MM(Cl) + 3*MM(O)
= 1*39.1 + 1*35.45 + 3*16.0
= 122.55 g/mol


mass(KClO3)= 25.5 g

number of mol of KClO3,
n = mass of KClO3/molar mass of KClO3
=(25.5 g)/(122.55 g/mol)
= 0.2081 mol
Balanced chemical equation is:
C12H22O11 + 8 KClO3 —> 12 CO2 + 11 H2O + 8 KCl


1 mol of C12H22O11 reacts with 8 mol of KClO3
for 0.0391 mol of C12H22O11, 0.3132 mol of KClO3 is required
But we have 0.2081 mol of KClO3

so, KClO3 is limiting reagent
Answer: KClO3 is limiting reagent

2)
we will use KClO3 in further calculation


Molar mass of KCl,
MM = 1*MM(K) + 1*MM(Cl)
= 1*39.1 + 1*35.45
= 74.55 g/mol

According to balanced equation
mol of KCl formed = (8/8)* moles of KClO3
= (8/8)*0.2081
= 0.2081 mol


mass of KCl = number of mol * molar mass
= 0.2081*74.55
= 15.5 g

Answer: 15.5 g

3)

According to balanced equation
mol of C12H22O11 reacted = (1/8)* moles of KClO3
= (1/8)*0.2081
= 0.026 mol
mol of C12H22O11 remaining = mol initially present - mol reacted
= 0.0391 - 0.026
= 0.0131 mol


Molar mass of C12H22O11,
MM = 12*MM(C) + 22*MM(H) + 11*MM(O)
= 12*12.01 + 22*1.008 + 11*16.0
= 342.296 g/mol

mass of C12H22O11,
m = number of mol * molar mass
= 1.314*10^-2 mol * 342.296 g/mol
= 4.50 g

Answer: 4.50 g

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