Respuesta :
Answer:
25.275 moles of oxygen gas will be required to completely react with all the samarium metal.
Explanation:
[tex]4Sm+3O_2\rightarrow 2Sm_2O_3[/tex]
Number of moles samarium metal = 33.7 moles
According to reaction, 4 moles of samarium reacts with 3 moles of oxygen gas.
Then 33.7 moles of samarium will react with:
[tex]\frac{3}{4}\times 33.7 mol=25.275 mol[/tex]of oxygen gas.
25.275 moles of oxygen gas will be required to completely react with all the samarium metal.
Answer:
Moles of oxygen gas required to react completely with 33.7 moles of samarium metal is [tex]\fbox{25.3 \text{ mol}}[/tex].
Explanation:
A chemical equation in which the number of atoms of each element is the same on the reactant and product side is called a balanced chemical equation.
The balanced chemical equation can be used to determine the stoichiometric ratio between the reactant and the product. The stoichiometric ratio thus enables us to calculate:
1. Amount of one reactant required to react completely with the other reactant.
2. Amount of the product that can be produced from the given amount of the reactant.
Step 1: Write the chemical equation for the reaction between samarium metal and oxygen gas.
The chemical formula for oxygen gas is [tex]\text{O}_{2}[/tex].
Samarium has +3 oxidation state within the samarium/oxygen compound. So, the chemical formula of the samarium oxygen compound is [tex]\text{Sm}_{2}\text{O}_{3}[/tex].
The chemical equation is as follows:
[tex]\fbox{\text{Sm}+\text{O}_{2} \rightarrow \text{Sm}_{2}\text{O}_{3}\\\end{minipage}}[/tex]
Step 2: Balance the chemical equation for the reaction between oxygen and samarium metal.
The number of oxygen atoms in the product side is 3 and in the reactant side is 2. Put coefficient 2 in front of [tex]\text{Sm}_{2}\text{O}_{3}[/tex] and 3 in front of [tex]\text{O}_{2}[/tex] to balance the oxygen atoms.
[tex]\fbox{\text{Sm}+3\text{O}_{2} \rightarrow 2 \text{Sm}_{2}\text{O}_{3}\\\end{minipage}}[/tex]
The number of samarium atoms in the product side is 4 and in the reactant side is 1. Put coefficient 4 in front of Sm in the reactant side.
[tex]\fbox{\\4\text{Sm}+3\text{O}_{2} \rightarrow 2 \text{Sm}_{2}\text{O}_{3}\\\end{minipage}}[/tex]
Step 3: Determine the stoichiometric ratio between samarium and oxygen from the above balanced chemical equation.
According to the balanced chemical equation, we can see that the stoichiometric ratio between samarium and oxygen is 4:3.
Step 4: Use unitary method and calculate the moles of oxygen required to completely react with the given moles of samarium metal as follows:
[tex]\text{moles of O}_{2} = \left( \text{moles of Sm} \right)\left( \frac{3 \text{ mol of O}_{2}}{4 \text{ mol of Sm}} \right)[/tex] ...... (1)
Step 5: Substitute 33.7 mol for moles of Sm in equation (1).
[tex]\text{Moles of O}_{2} = \left( \text{33.7 mol} \right) \left( \frac{3 \text{ mol of O}_{2}}{4 \text{ mol of Sm}} \right)\\\text{Moles of O}_{2}= 25.275 \text{ mol}\\\text{Moles of O}_{2}= 25.3 \text{ mol}[/tex]
Note:
Do not forgot to balance the reaction. The reaction must be balanced in order to calculate the amount (mol) of oxygen required to completely react with the given amount of samarium.
Learn more:
1. Balanced chemical equation https://brainly.com/question/1405182
2. Learn more about how to calculate moles of the base in given volume https://brainly.com/question/4283309
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Some basic concept of chemistry
Keywords: samarium, oxygen gas, samarium/oxygen compound, 33.7 moles, 25.3 mol, balanced equation, stoichiometric ratio, coefficient, balance, moles, completely react.
