1. Silver nitrate will react with aluminum metal, yielding aluminum nitrate and silver metal. If you start with 0.223 moles of aluminum, you will create __ grams of silver.

__ AgNO3 + __ Al ⟶__ Ag + __ Al(NO3)3

2. Mercury combines with oxygen to yield mercury (II) oxide. If you want to produce 3.12 moles of mercury (II) oxide, you will need to start with grams of mercury.

__ Hg + __O2 ⟶__ HgO

3. Nitrogen and oxygen combine to produce dinitrogen pentoxide. If you want to produce 12.99 moles of dinitrogen pentoxide, you need ___ grams of oxygen available.

__ N2 + __ O2 ⟶__ N2O5

4. Benzene is an organic solvent that releases a lot of energy when it is burned. Burning 0.103 moles of benzene will release ___ grams of carbon dioxide.

__ C6H6 + __ O2 ⟶ __ CO2 + __ H2O

Explanation:

1)

Given data:

Number of moles of aluminium = 0.223 mol

Mass of silver produced = ?

Solution:

Chemical equation:

3AgNO₃  +   Al  →  3Ag + Al(NO₃)₃

Now we will compare the moles of Al with silver.

                               Al           :            Ag

                                1            :             3

                                0.223   :         3×0.223= 0.669 mol

Grams of silver:

Mass = number of moles × molar mass

Mass = 0.669 mol × 107.87 g/mol

Mass = 72.2 g

2)

Given data:

Number of moles of mercury(II) oxide produced = 3.12 mol

Mass of mercury = ?

Solution:

Chemical equation:

2Hg + O₂  →   2HgO

Now we will compare the moles of mercury with mercury(II) oxide.

                         HgO         :         Hg

                            2            :          2

                          3.12          :       3.12

Mass of Hg:

Mass = number of moles × molar mass

Mass = 3.12 mol × 200.59 g/mol

Mass = 625.84 g

3)

Given data:

Number of moles of dinitrogen pentoxide = 12.99 mol

Mass of oxygen = ?

Solution:

Chemical equation:

2N₂  + 5O₂   →  2N₂O₅

Now we will compare the moles of N₂O₅ with oxygen.

                 N₂O₅          :           O₂

                     2            :             5

                    12.99      :         5/2×12.99 = 32.48 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 32.48 mol × 32 g/mol

Mass = 1039.36 g

4)

Given data:

Number of moles of benzene = 0.103 mol

Mass of carbon dioxide = ?

Solution:

Chemical equation:

2C₆H₆  + 15O₂   →  12CO₂ + 6H₂O

Now we will compare the moles of N₂O₅ with oxygen.

                  C₆H₆         :           CO₂

                     2            :             12

                    0.103      :         12/2×0.103 = 0.618 mol

Mass of carbon dioxide:

Mass = number of moles × molar mass

Mass = 0.618 mol × 44 g/mol

Mass = 27.192 g