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Nonamiya [84]
3 years ago
13

What mass of TiCl4 must react with an excess of water to produce 50.0g of TiO2 if the reaction has a 78.9% yield

Chemistry
1 answer:
Kipish [7]3 years ago
5 0

Answer:

\large \boxed{\text{150 g TiCl}_{4}}  

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:     189.68                  79.87

           TiCl₄ + 2H₂O ⟶ TiO₂ + 4HCl

m/g:                                 50.0

To solve this stoichiometry problem, you must

  • Convert the actual yield to the theoretical yield  
  • Use the molar mass of TiO₂ to convert the theoretical yield of TiO₂ to moles of TiO₂
  • Use the molar ratio to convert moles of TiO₂ to moles of TiCl₄
  • Use the molar mass of TiCl₄ to convert moles of TiCl₄ to mass of TiCt₄

1. Theoretical yield of TiO₂

\text{Theoretical yield} = \text{50.0 g actual} \times \dfrac{\text{100 g theoretical}}{\text{78.9 g actual}} = \text{63.37 g theoretical}

2.  Moles of TiO₂

\text{Mass of TiO}_{2} = \text{63.37 g TiO}_{2} \times \dfrac{\text{1 mol TiO}_{2}}{\text{79.87 g TiO}_{2} } = \text{0.7934 mol TiO}_{2}

3,  Moles of TiCl₄

The molar ratio is 1 mol TiO₂:1 mol TiCl₄.

\text{Moles of TiCl}_{4} = \text{0.7934 mol TiO}_{2} \times \dfrac{\text{1 mol TiCl}_{4}}{\text{1 mol TiO}_{2}} = \text{0.7934 mol TiCl}_{4}

4.  Mass of TiCl₄

\text{Mass of TiCl}_{4} = \text{0.7934 mol TiCl}_{4} \times \dfrac{\text{189.98 g TiCl}_{4}}{\text{1 mol TiCl}_{4}} =\textbf{150 g TiCl}_{\mathbf{4}} \\\\\text{You must use $\large \boxed{\textbf{150 g TiCl}_{\mathbf{4}}}$}

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romanna [79]

Answer :

(a) The limiting reactant is, O_2

(b) The mass of excess reactant is, 1.7646 g

(c) The theoretical yield of NO is, 2.616 g

(d) The percent yield of the reaction is, 37.46 %

Explanation : Given,

Mass of NH_3 = 3.25 g

Mass of O_2 = 3.50 g

Molar mass of NH_3 = 17 g/mole

Molar mass of O_2 = 32 g/mole

Molar mass of NO = 30 g/mole

First we have to calculate the moles of NH_3 and O_2.

\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}=\frac{3.25g}{17g/mole}=0.191moles

\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{3.50g}{32g/mole}=0.109moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

4NH_3+5O_2\rightarrow 4NO+6H_2O

From the balanced reaction we conclude that

As, 5 moles of O_2 react with 4 mole of NH_3

So, 0.109 moles of O_2 react with \frac{4}{5}\times 0.109=0.0872 moles of NH_3

From this we conclude that, NH_3 is an excess reagent because the given moles are greater than the required moles and O_2 is a limiting reagent and it limits the formation of product.

The limiting reactant is, O_2

Excess moles of NH_3 = 0.191 - 0.0872 = 0.1038 mole

Now we have to calculate the mass of NH_3.

\text{Mass of }NH_3=\text{Moles of }NH_3\times \text{Molar mass of }NH_3

\text{Mass of }NH_3=(0.1038mole)\times (17g/mole)=1.7646g

The mass of excess reactant = 1.7646 g

Now we have to calculate the moles of NO.

As, 5 moles of O_2 react with 4 mole of NO

So, 0.109 moles of O_2 react with \frac{4}{5}\times 0.109=0.0872 moles of NO

Now we have to calculate the mass of NO.

\text{Mass of }NO=\text{Moles of }NO\times \text{Molar mass of }NO

\text{Mass of }NO=(0.0872mole)\times (30g/mole)=2.616g

Now we have to calculate the percent yield of NO

\%\text{ yield of }NO=\frac{\text{Actual yield of }NO}{\text{Theoretical yield of }NO}\times 100=\frac{0.98g}{2.616g}\times 100=37.46\%

The percent yield of the reaction is, 37.46 %

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Divide 1.25/3.93 and then multiply by 100 to get the percent. This equals 31.81%
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Answer:

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In the Ka expression, we have a<u> proportional relationship</u> between Ka and the concentration of H^+. Therefore, if we have a higher Ka value we will have a smaller pH (lets keep in mind that with a higher

So, if we have to find the higher pH value we need to search the <u>smaller Ka value</u> in this case HCN~~Ka=4.9x10^-^1^0.

I hope helps!

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