Question

reaction amounts
substance amount (reacted or formed)
kclo₃ 10.0 g
kcl 7.5 g
al 9.0 g
reaction 1: kclo₃(s) → kcl(s) + o₂(g)
reaction 2: al(s) + o₂(g) → al₂o₃(s)
the decomposition of potassium chlorate in reaction 1 provides the necessary oxygen for reaction 2 to take place.
determine the coefficients that will balance reactions 1 & 2. also find the mass of al₂o₃ that will form given the amounts in the table.
a reaction 1: 2, 2, and 3
reaction 2: 2, 3, and 2
7.5 grams al₂o₃
b reaction 1: 1, 1, and 2
reaction 2: 2, 3, and 2
9.5 grams al₂o₃
c reaction 1: 2, 2, and 3
reaction 2: 4, 3, and 2
11.5 grams al₂o₃
d reaction 1: 2, 2, and 3
reaction 2: 2, 2, and 3
16.5 grams al₂o₃

Explanation:

Step1: Balance Reaction 1

The balanced equation for the decomposition of potassium chlorate ($KClO_3$) is $2KClO_3(s)
ightarrow 2KCl(s)+3O_2(g)$. This is achieved by ensuring the number of potassium (K), chlorine (Cl), and oxygen (O) atoms are the same on both sides of the equation.

Step2: Balance Reaction 2

The balanced equation for the reaction of aluminum (Al) with oxygen ($O_2$) is $4Al(s)+3O_2(g)
ightarrow 2Al_2O_3(s)$. We balance the Al and O atoms on both sides.

Step3: Calculate moles of Al

The molar - mass of Al is $M_{Al}=26.98\ g/mol$. The number of moles of Al, $n_{Al}=\frac{m_{Al}}{M_{Al}}=\frac{9.0\ g}{26.98\ g/mol}\approx0.333\ mol$.

Step4: Calculate moles of $Al_2O_3$ from moles of Al

From the balanced equation of Reaction 2, the mole - ratio of Al to $Al_2O_3$ is 4:2 or 2:1. So, the number of moles of $Al_2O_3$, $n_{Al_2O_3}=\frac{1}{2}n_{Al}=\frac{1}{2}\times0.333\ mol = 0.1665\ mol$.

Step5: Calculate mass of $Al_2O_3$

The molar - mass of $Al_2O_3$ is $M_{Al_2O_3}=2\times26.98\ g/mol + 3\times16.00\ g/mol=101.96\ g/mol$. The mass of $Al_2O_3$, $m_{Al_2O_3}=n_{Al_2O_3}\times M_{Al_2O_3}=0.1665\ mol\times101.96\ g/mol\approx16.5\ g$.

Answer:

D. Reaction 1: 2, 2, and 3; Reaction 2: 2, 2, and 3; 16.5 grams $Al_2O_3$