Answer:
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Explanation:
If 36g of Al react with 34L of Oz, 0.667 moles of Al2O3 are produced. The correct option is b, 0.667 moles
The mole is a SI unit of measurement that is used to calculate the quantity of any substance.
[tex]\rm 4Al +3O_2 -- > 2Al_2O_3[/tex]
Step1- calculate the moles of aluminum
The mass of Al is 36 g
The molar mass of Al is 27
[tex]\rm Number\;of \;moles= \dfrac{mass}{molar\;mass}\\\rm Number\;of \;moles\;of\;Al = \dfrac{36}{27} = 1.33\;mol[/tex]
Step2- calculate the moles of oxygen
The mass of O₂ is 34 L
[tex]\rm 1 L = \dfrac{1}{22.4 }[/tex]
[tex]\rm Number\;of \;moles\;O_2= \dfrac{34}{22.4} = 1.51[/tex]
The limiting reagent is 1.33 MOL. By the rule of stoichiometry, four moles of Al produced 2 moles of Al.
[tex]\rm \dfrac{1.33}{2} = 0.665 \;moles[/tex]
Thus, the correct option is b. 0.667 moles.
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