Question

1. Sodium and water react according to the following equation. If 31.5g of sodium are added to excess water, how many liters of hydrogen gas are formed at STP?
2Na+2H2O—>2NaOH+H2

2. Sodium reacts with oxygen to produce sodium oxide as described by the balanced equation below. If 54.1g of sodium reacts with excess oxygen gas to produce 61.8g Of sodium oxide, what is the percent yield? (Hint: be sure to calculate theoretical yield first)
4Na+O2–>2Na2O

(Please show work)

Answer 1

 Question  1

 The number  of liter   of  hydrogen gas  that are formed   at STP is  

15.344 L

 calculation

2 Na   +2 H₂O →  2NaOH  +H₂

 calculate  the  moles of   Na

moles  =mass/molar mass

from periodic table the molar mass of Na =  23 g/mol

moles = 31.5 g /23 g /mol  = 1.37  moles

From  equation above the   mole   ratio  of  Na: H2   is  2:1  

therefore the  moles of H2  =1.37 x1/2 =0.685  moles

At  STP  1  moles  of a gas  = 22.4 L

              0.685   moles  =  ?  L

by cross multiplication

=[ (0.685  moles  x 22.4 L) /  1  mole  =15.344 L

Question 2

The  percent   yield       = 84.77%

calculation

4 Na + O₂   →   2 Na₂O

%yield  =actual yield / theoretical  yield  x 100

Actual  yield=  61.8 g

 The theoretical  yield  is calculated as  below

Step  1:  find the  moles  of Na

     moles =  mass /molar mass

 from periodic table the  molar mass  of Na = 23 g/mol

moles = 54.1 g /23 g/mol =2.352  moles

Step 2: use the  mole ratio to determine  the  moles  of Na₂O

 Na:Na₂O  is  4: 2  therefore the moles of Na₂O = 2.352  x 2/4 =1.176 moles

Step 3:   find  the theoretical mass  of Na₂O

mass =  moles  x  molar  mass

The molar mass  of Na₂O =  [(23 x 2 ) + 16]  =62 g/mol

mass  = 1.176 moles   x 62 g/mol = 72.9  g

%   yield   is therefore  = 61.8 g /72.9 x 100  = 84.77%