reaction amounts\nsubstance amount (reacted or formed)\nkclo₃ 10.0 g\nkcl 7.5 g\nal 9.0 g\nreaction 1…

reaction amounts\nsubstance amount (reacted or formed)\nkclo₃ 10.0 g\nkcl 7.5 g\nal 9.0 g\nreaction 1: kclo₃(s) → kcl(s) + o₂(g)\nreaction 2: al(s) + o₂(g) → al₂o₃(s)\nthe decomposition of potassium chlorate in reaction 1 provides the necessary oxygen for reaction 2 to take place.\ndetermine the coefficients that will balance reactions 1 & 2. also find the mass of al₂o₃ that will form given the amounts in the table.\na reaction 1: 2, 2, and 3\nreaction 2: 2, 3, and 2\n7.5 grams al₂o₃\nb reaction 1: 1, 1, and 2\nreaction 2: 2, 3, and 2\n9.5 grams al₂o₃\nc reaction 1: 2, 2, and 3\nreaction 2: 4, 3, and 2\n11.5 grams al₂o₃\nd reaction 1: 2, 2, and 3\nreaction 2: 2, 2, and 3\n16.5 grams al₂o₃

reaction amounts\nsubstance amount (reacted or formed)\nkclo₃ 10.0 g\nkcl 7.5 g\nal 9.0 g\nreaction 1: kclo₃(s) → kcl(s) + o₂(g)\nreaction 2: al(s) + o₂(g) → al₂o₃(s)\nthe decomposition of potassium chlorate in reaction 1 provides the necessary oxygen for reaction 2 to take place.\ndetermine the coefficients that will balance reactions 1 & 2. also find the mass of al₂o₃ that will form given the amounts in the table.\na reaction 1: 2, 2, and 3\nreaction 2: 2, 3, and 2\n7.5 grams al₂o₃\nb reaction 1: 1, 1, and 2\nreaction 2: 2, 3, and 2\n9.5 grams al₂o₃\nc reaction 1: 2, 2, and 3\nreaction 2: 4, 3, and 2\n11.5 grams al₂o₃\nd reaction 1: 2, 2, and 3\nreaction 2: 2, 2, and 3\n16.5 grams al₂o₃

Answer

Explanation:

Step1: Balance Reaction 1

The balanced equation for the decomposition of potassium chlorate ($KClO_3$) is $2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g)$. 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 $4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)$. 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 $Al_2O_3$