Answer to Question #349845 in Inorganic Chemistry for MAT

Question

A. 200.0 g of Ca(C₂H₃O₂)₂ was found in the chemistry laboratory. 

      Calculate

          i. the molecular mass of 1 mole Ca(C₂H₃O₂) _2.

         ii. the number of  moles of Ca(C₂H₃O₂)₂  that exists   

        iii. the number of Ca(C₂H₃O₂)₂ molecules present

        iv.  the volume at STP  if Ca(C₂H₃O₂)₂ is converted to gas.

        v.  the molarity of a solution made by dissolving 200.0 g of      

                Ca(C₂H₃O₂)₂ in 200 ml of water

b)  Answer the following stoichiometry questions by referring to the equation below:

2 KClO₃    2 KCl + 3 O₂

i.        If 1.50 moles of KClO₃ decomposes, what is the mass of O₂ that will be produced?

ii.       If 80.0 grams of O₂ was produced, how many moles of KClO₃                  are decomposed?

iii.    Find the mass of KClO₃ needed if we need to produce 2.75                                 moles of KCl.

Answer 1

A.)

i.

M_r [Ca(C₂H₃O₂)₂] = 40 + (12*2 + 1*3 + 16*2)* 2 = 158 g/mol

ii.

n = m / M_r = 200 g / 158 g/mol = 1,266 mol

iii.

1 mol => 6,02*10²³ molecules

1,266 mol => X

X = 1,266 * 6,02*10²³/1 = 7,62*10²³

iv.

1 mol gas = 22,4 liters

1,266 mol = X

X = 1,266 * 22,4 / 1 = 28,36 liters

v.

C_M = n / V = 1,266 mol / 0,2 liters = 6,33 M

B.)

i.

2KClO₃ => 2KCl + 3O₂

2mol => 96 g

1,5mol => X

X = 1,5 * 96 / 2 = 72 g

ii.

2KClO₃ => 2KCl + 3O₂

2mol => 96 g

X => 80 g

X = 2 * 80 / 96 = 1,667 mol

iii.

KClO₃ => KCl + 1,5 O2

120,5 g => 1 mol

X => 2,75 mol

X = 120,5 * 2,75 / 1 = 331,375 g