what is the limiting reactant when 7.0 grams of aluminum and 15.2 grams of oxygen react to produce aluminum oxide?

The number of moles of Al₂O₃ produced by Al are less it will limiting reactant.

Explanation:

Given data:

Mass of aluminium = 7.0 g

Mass of oxygen = 15.2 g

Limiting reactant = ?

Solution:

Chemical equation:

4Al + 3O₂ →  2Al₂O₃

Number of moles of Aluminium:

Number of moles = mass/ molar mass

Number of moles = 7.0 g / 27 g/mol

Number of moles = 0.3 mol

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 15.2 g / 32 g/mol

Number of moles = 0.5 mol

Now we will compare the moles of aluminium oxide with oxygen and Al.

                         O₂             :         Al₂O₃

                          3              :              2

                         0.5            ;            2/3×0.5 = 0.33 mol

                         Al              :               Al₂O₃

                          4              :                    2

                          0.3           :              2/4×0.3 = 0.15 mol

The number of moles of Al₂O₃ produced by Al are less it will limiting reactant.

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1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation: