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 ( show work)

 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%

For cyclohexane compounds drawn in a chair conformation:   -axial bonds are vertical  -equatorial bonds are parallel to two bonds within the ring.  thus, as i assume the chair conformation is drawn:   -the x group is in an axial position, as are the five other vertical bonds.  -three axial bonds are on the top face of the ring  -the other three axial bonds are on the bottom face  the position of the second substituent on each ring carbon is equatorial.  gauche groups must be on an adjacent carbon such as carbon 6 and carbon 2, and also gauche groups are rotated by 60 degrees relative to each other.  drawing the newman projections for c1–c2 and c1–c6 aids in visualizing the groups that are gauche to the y group on c1.  click this link and watch some academic videos that may you more understand the topic 

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