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4 years ago

Which expression could represent the

1 answer:

6 0

There are a couple of ways in which you can express the concentration of a solution, and here they are: gram per liter (g/L), molarity (M), parts per million (ppm.), and percents (%).
As you can see, only M appears in your answers, which means that the correct option should be (2) 3.5 M.

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Read 2 more answers

Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide: 4Al (s) + 3O2 (g) â†’ 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 %

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