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Marat540 [252]
3 years ago
14

5. A vitamin contains 1.5 x 10' grams of zinc. How many moles of zinc are in the vitamin?

Chemistry
1 answer:
Anni [7]3 years ago
7 0

Answer:

                      2.29 × 10⁻⁴ moles

Explanation:

Data Given:

                 Chemical Symbol of Zinc  =  Zn

                 Mass of Zinc   =  1.5 × 10⁻² g

                 A.Mass of Zinc  =  65.38 g.mol⁻¹

The Atomic Mass of Zinc can be obtained from periodic table. While, the mass is given in statement. Also, it is important to know the symbol because some elements exist in diatomic or polyatomic forms i.e. H₂, N₂, S₈ e.t.c.

                 Mole is the unit used to calculate the amount of substance. It has following general forms,

                                             Mole =  Mass / M.Mass

And,

                     Mole =  # of Particles / 6.022 × 10²³ particles.mol⁻¹

Hence, using first equation,

                     Moles  =  1.5 × 10⁻² g / 65.38 g.mol⁻¹

                     Moles =  2.29 × 10⁻⁴ moles

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<u>Answer:</u> The percentage yield of HF is 73.36 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     ....(1)  

For calcium fluoride:

Given mass of calcium fluoride = 6.25 kg  = 6250 g   (Conversion factor: 1 kg = 1000 g)

Molar mass of calcium fluoride = 78.07 g/mol

Putting values in above equation, we get:  

\text{Moles of calcium fluoride}=\frac{6250g}{78.07g/mol}=80.05mol

For the given chemical reaction:

CaF_2+H_2SO_4\rightarrow CaSO_4+2HF

By Stoichiometry of the reaction:

1 mole of calcium fluoride produces 2 moles of hydrofluoric acid

So, 80.05 moles of calcium fluoride will produce = \frac{2}{1}\times 80.05=160.1mol of hydrofluoric acid

Now, calculating the theoretical yield of hydrofluoric acid using equation 1, we get:

Moles of of hydrofluoric acid = 160.1 moles

Molar mass of hydrofluoric acid = 20.01 g/mol

Putting values in equation 1, we get:

160.1mol=\frac{\text{Theoretical yield of hydrofluoric acid}}{20.01g/mol}=3203.6g=3.20kg

To calculate the percentage yield of hydrofluoric acid, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of hydrofluoric acid = 2.35 kg

Theoretical yield of hydrofluoric acid = 3.20 kg

Putting values in above equation, we get:

\%\text{ yield of hydrofluoric acid}=\frac{2.35g}{3.20g}\times 100\\\\\% \text{yield of hydrofluoric acid}=73.36\%

Hence, the percentage yield of HF is 73.36 %

4 0
3 years ago
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