CHEMISTRY HIGH SCHOOL

How many moles of atoms are in 2.00 g of 13c?

Answers

Answer 1
Answer:

There are 0.154 moles of atoms in 2.00 g of carbon-13.

Further Explanation:

Moles, Atomic mass and Molecular mass  

  • 1 mole of a pure substance contains a mass that is equal to the relative atomic mass or molecular mass of the substance.
  • Therefore; molar mass is given as grams per mole of a substance  

Hence;

Molar mass = mass of a substance/ Number of moles  

     g/mol = g /mole  

Thus;

  • M = m/n; where M is the molar mass, m is the mass and n is the number of moles
  • From this relationship we can therefore, calculate mass by multiplying the number of moles by molar mass of a substance.

That is; Mass = moles x molar mass  

  • To calculate number of moles;

We have; n = m/M

Number of moles = Mass of the substance/ Molar mass

  • In our case;

Mass = 2 g of Carbon-13  

Molar mass = 13.0 g/mol  

Since; Number of moles = Mass/ molar mass

Thus;

Moles = (2.0 g)/ (13.0 g/mol)

          = 0.154 moles  

Keywords: Moles, Molecular mass, relative atomic mass  

Learn more about:

Level: High school  

Subject: Chemistry  

Topic: Moles  

Sub-topic: Moles, molecular mass and mass of a pure substance.

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The following shows the precipitation reaction of barium chloride (BaCl₂) and sodium hydroxide (NaOH): BaCl₂(aq) + NaOH(aq) ----> Ba(OH)₂(s) + 2NaCl(aq)
If you have to prepare the reactants by dissolving 1.00 g of BaCl₂ and 1.00 g of NaOH in water,
(a) What is the limiting reactant?
(b) How many grams of Ba(OH)₂ are produced?
(c) If your experiment produced 0.700 g of Ba(OH)₂, what is the percent yield of Ba(OH)₂?
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Answers

Answer:

The answer to your question is:

a) BaCl2

b) 0.8208 g

c) yield = 85.3 %

d)

Explanation:

                     BaCl₂(aq) + NaOH(aq) ----> Ba(OH)₂(s) + 2NaCl(aq)

Data

a) 1 g of BaCl₂

   1 g of NaOH

MW BaCl2 = 137 + (35.5x2) = 208 g

MW NaOH = 23 + 16 + 1 = 40 g

                               208 g of BaCl2 -------------  1 mol

                                    1 g of BaCl2 -------------    x

                                  x = ( 1 x 1) / 208 = 0.0048 mol of BaCl2

                                    40 g of NaOH ------------  1 mol

                                       1 g of NaOH ------------   x

                                 x = (1 x 1) / 40

                                 x = 0.025 mol of NaOH

The ratio BaCl2 to NaOH is 1:1 (in the equation)

But experimentally we have 0.0048 : 0.025, so the limiting reactant is BaCl2, because is in lower concentration.

b)

                    1 mol of BaCl2 -------------- 1 mol of Ba(OH)2

                    0.0048 mol     ---------------   x

                     x = (0.0048 x 1) / 1

                     x = 0.0048 mol of Ba(OH)2

MW Ba(OH)2 = 137 + 32 + 2 = 171 g

                     171 g of Ba(OH)2 -------------------- 1 mol

                      x                         --------------------  0.0048 mol

                     x = (0.0048 x 171) / 1

                     x = 0.8208 g

c)Data

Ba(OH)2 = 0.700 g

                   % yield = 0.700 / 0.8208 x 100

                   % yield = 85.3

d)

Sorry, i don't understand this question

       

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Answers

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Explanation:

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