6. Tristan completed a lab that combusted 5.0 grams of sodium
Write a complete balanced equation for the combustion of sodium.
a
b. What is the maximum amount of product that Tristan can obtain in this lab?

4Na  +   O₂  →   2Na₂O  

6.82g

Explanation:

The combustion of sodium involves combining the element with oxygen gas:

              4Na  +   O₂  →   2Na₂O  

We have been given 5g of Na.

The balanced reaction equation is given as shown above. This reaction is also a synthesis or combination reaction.

b. Maximum amount of product that can be obtained.

Sodium is the limiting reactant and it will determine the amount of product that can be formed.

  To solve this, we find the number of moles of the given sodium.

  Mass of given sodium  = 5g

    Molar mass of Na  = 23g/mol

  Number of moles of Na  =  

Insert the parameters and solve;

 Number of moles of Na  =    = 0.217moles

  From the balanced reaction equation:

         4 moles of Na will produce 2 moles of Na₂O  

      0.217 moles of Na will produce   = 0.11mole of Na₂O  

So;

 Mass of Na₂O   = number of moles x molar mass

    Molar mass of Na₂O   = 2(23) + 16  = 62g/mol

 Mass of Na₂O   = 0.11 x 62  = 6.82g

93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.