Aluminum and cloride undergoes a synthesis reaction if 8molnof Al reacts with 10mol of cl, what is the maximum amount of AlCl3 can produce

Question

Question

in progress 0

Chemistry Thu Giang 4 years 2021-07-19T14:55:34+00:00 2021-07-19T14:55:34+00:00 1 Answers 35 views 0

Answers ( )

Name*

E-Mail*

Website

Featured image

Browse

Captcha* Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )

Answer*

bonexptip · Answer 1 · 2021-07-19T14:57:01+00:00

Answer: The mass of $AlCl_3$ produced is 889.38 g

Explanation:

We are given:

Moles of Al = 8 mol

Moles of $Cl_2$ = 10 mol

For the given chemical reaction:

$2Al+3Cl_2\rightarrow 2AlCl_3$

By stoichiometry of the reaction:

If 3 moles of chlorine gas reacts with 2 moles of Al

So, 10 moles of chlorine gas will react with = $\frac{2}{3}\times 10=6.67mol$ of Al

As the given amount of Al is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, chlorine gas is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 3 moles of $Cl_2$ produces 2 mole of $AlCl_3$

So, 10 moles of $Cl_2$ will produce = $\frac{2}{3}\times 10=6.67mol$ of $AlCl_3$

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

We know, molar mass of $AlCl_3$ = 133.34 g/mol

Putting values in above equation, we get:

$\text{Mass of }AlCl_3=(6.67mol\times 133.34g/mol)=889.38g$

Hence, the mass of $AlCl_3$ produced is 889.38 g