1. the space shuttle used two solid rocket boosters to launch it into space. the reaction that occurred is…

1. the space shuttle used two solid rocket boosters to launch it into space. the reaction that occurred is written below: nh₄clo₄(s) + al(s) --> no(g) + al₂o₃(s) + h₂o(g) + alcl₃(g)\na. balance the reaction and rewrite.\nb. how many grams of aluminum would be needed to react with 56 kg of ammonium perchlorate?\nc. if 67 grams of ammonium perchlorate were to react with 10 grams of aluminum:\na. which material would be the limiting reactant?\nb. what would be the theoretical yield of nitrogen monoxide?\nd. was the aluminum oxidized or reduced in the reaction?

1. the space shuttle used two solid rocket boosters to launch it into space. the reaction that occurred is written below: nh₄clo₄(s) + al(s) --> no(g) + al₂o₃(s) + h₂o(g) + alcl₃(g)\na. balance the reaction and rewrite.\nb. how many grams of aluminum would be needed to react with 56 kg of ammonium perchlorate?\nc. if 67 grams of ammonium perchlorate were to react with 10 grams of aluminum:\na. which material would be the limiting reactant?\nb. what would be the theoretical yield of nitrogen monoxide?\nd. was the aluminum oxidized or reduced in the reaction?

Answer

Explanation:

Step1: Balance the chemical equation

We balance the equation $NH_4ClO_4(s)+Al(s)\rightarrow NO(g)+Al_2O_3(s)+H_2O(g)+AlCl_3(g)$ by adjusting the coefficients. The balanced equation is $3NH_4ClO_4(s)+3Al(s)\rightarrow 3NO(g)+Al_2O_3(s)+6H_2O(g)+AlCl_3(g)$.

Step2: Calculate moles of ammonium perchlorate in part B

First, convert the mass of ammonium perchlorate to moles. The molar - mass of $NH_4ClO_4$ is $M_{NH_4ClO_4}=14 + 4\times1+35.5 + 4\times16=117.5\ g/mol$. Given $m = 56\ kg=56000\ g$, the number of moles $n_{NH_4ClO_4}=\frac{m}{M}=\frac{56000\ g}{117.5\ g/mol}\approx476.6\ mol$. From the balanced equation, the mole - ratio of $NH_4ClO_4$ to $Al$ is $1:1$. So the moles of $Al$ needed $n_{Al}=n_{NH_4ClO_4}\approx476.6\ mol$. The molar - mass of $Al$ is $M_{Al}=27\ g/mol$. The mass of $Al$ needed $m_{Al}=n_{Al}\times M_{Al}=476.6\ mol\times27\ g/mol = 12868.2\ g$.

Step3: Determine limiting reactant in part C - a

The molar - mass of $NH_4ClO_4$ is $117.5\ g/mol$ and the molar - mass of $Al$ is $27\ g/mol$. For $67\ g$ of $NH_4ClO_4$, the number of moles $n_{NH_4ClO_4}=\frac{67\ g}{117.5\ g/mol}\approx0.57\ mol$. For $10\ g$ of $Al$, the number of moles $n_{Al}=\frac{10\ g}{27\ g/mol}\approx0.37\ mol$. From the balanced equation, the mole - ratio of $NH_4ClO_4$ to $Al$ is $1:1$. Since $0.37\ mol$ of $Al$ is less than $0.57\ mol$ of $NH_4ClO_4$, $Al$ is the limiting reactant.

Step4: Calculate theoretical yield of NO in part C - b

From the balanced equation, the mole - ratio of $Al$ to $NO$ is $1:1$. Since $n_{Al}=0.37\ mol$, $n_{NO}=0.37\ mol$. The molar - mass of $NO$ is $M_{NO}=14 + 16=30\ g/mol$. The theoretical yield of $NO$ is $m_{NO}=n_{NO}\times M_{NO}=0.37\ mol\times30\ g/mol = 11.1\ g$.

Step5: Determine oxidation - reduction in part D

In the reaction, the oxidation state of $Al$ changes from $0$ in $Al(s)$ to $+ 3$ in $Al_2O_3$ and $AlCl_3$. Since the oxidation state of $Al$ increases, $Al$ is oxidized.

Answer:

A. $3NH_4ClO_4(s)+3Al(s)\rightarrow 3NO(g)+Al_2O_3(s)+6H_2O(g)+AlCl_3(g)$ B. $12868.2\ g$ C. a. $Al$ is the limiting reactant b. $11.1\ g$ D. $Al$ is oxidized