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Butoxors [25]
4 years ago
6

A chemist adds of a M mercury(II) iodide solution to a reaction flask. Calculate the mass in milligrams of mercury(II) iodide th

e chemist has added to the flask. Round your answer to significant digits.
Chemistry
1 answer:
Oksi-84 [34.3K]4 years ago
7 0

Answer:

0.42 mg

Explanation:

It seems the question is incomplete, however I'll use values that have been found in a web search:

"A chemist adds 55.0 mL of a 1.7x10⁻⁵ M mercury(II) iodide solution to a reaction flask. Calculate the mass in milligrams of mercury(II) iodide the chemist has added to the flask. Round your answer to 2 significant digits."

<u>Keep in mind that while the process of solving the problem remains the same, if the values in your question are different, your answer will differ as well</u>.

First we <u>calculate the moles of mercury (II) iodide</u> (HgI₂):

  • 1.7x10⁻⁵M * 55.0 mL = 9.35x10⁻⁴ mmol HgI₂

Then we<u> convert mmol to mg, using the molar mass</u> (454.4 g/mol):

  • 9.35x10⁻⁴ mmol HgI₂* 454.4 mg/mmol = 0.42 mg
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Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide: 4Al (s) + 3O2 (g) → 2Al2O3 (s) In
jek_recluse [69]

Answer: The percent yield of the reaction is 74 %

Explanation:

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

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

\text{Moles of aluminium}=\frac{2.5g}{27g/mol}=0.092mol

For oxygen gas:

\text{Moles of oxygen gas}=\frac{2.5g}{32g/mol}=0.078mol

The chemical equation for the reaction of titanium and chlorine gas follows:

4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)

By Stoichiometry of the reaction:

4 moles of aluminium reacts with 3 moles of oxygen.

So, 0.092 moles of aluminium reacts with = \frac{3}{4}\times 0.092=0.069mol of oxygen

As, given amount of oxygen is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

4 moles of aluminium produce = 2 moles of Al_2O_3

So, 0.092 moles of aluminium will produce = \frac{2}{4}\times 0.092=0.046moles of Al_2O_3

Now, calculating the mass of aluminium oxide:

\text{Mass of aluminium oxide}=moles\times {\text {molar mas}}=0.046mol\times 102g/mol=4.7g

To calculate the percentage yield of titanium (IV) chloride, we use the equation:

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

Experimental yield  = 3.5 g

Theoretical yield = 4.7 g

Putting values in above equation, we get:

\%\text{ yield of reaction}=\frac{3.5g}{4.7g}\times 100\\\\\% \text{yield of reaction}=74\%

Hence, the percent yield of the reaction is 74 %

6 0
3 years ago
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