Based on the following equation: $2 Al + 3 CuCl_2 -> 2AlCl_3 + 3 Cu$. lf 10.53 grams of $CuCl_2$ were reacted, how many grams of $AlCl_3$ will be produced?

Md Ariful Islam

0 like 0 dislike
1 views
Share with your friends

1 Answers

Md Ariful Islam

Call

According to this law a reactant will produce same no of equivalent of product/s as its reacting mass has.
here the equivalent mass of $CuCl_2$ = Formula wt /total valency of metal(Cu) = $(63.55+2*35.55)/(1*2)$ =67.33
Again
the equivalent mass of $AlCl_3$ = Formula wt /total valency of metal(Al) = $(27+3*35.55)/(1*3)$ =44.55
Now no. of gram equivalent in 10.53g of $CuCl_2$ reacted = $10.53/67.33$=0.156
hence the same no.of gram equivalent of $AlCl_3$ will be produced
So the mass of $AlCl_3$ produced =0.156x equivalent mass of $AlCl_3$ 0.156x44.55 = 6.95 g

Pl, give feed back

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
In the reaction: $A + B -> C$, if 4.2 grams of A react to make 6.8 grams of C, then how many grams of B were reacted?

We can use the Law of Conservation of Mass to solve the problem. $underbrace("A")_color(red)("4.2 g") + underbrace("B")_color(red)(x) → underbrace("C")_color(red)("6.8 g")$ $"4.2 g"color(white)(l) + x = "6.8 g"$ $x = "6.8...

1 Answers 1 views
What is the reaction between $CuCl_2$, and aluminum metal to give copper metal and $AlCl_3$?

$Al rarr Al^(3+) + 3e^-$; oxidation. $CuCl_2(aq) +2e^(-) rarr Cu(s)darr + 2Cl^-$; reduction. We cross-mulitply to remove the electrons. $3CuCl_2(aq) +2Al(s) rarr 2AlCl_3(aq) + 3Cu(s)darr$ A sheet of aluminum place...

1 Answers 1 views
How do you balance $Al + CuCl_2 -> AlCl_3 + Cu$?

Aluminum is oxidized to $Al^(3+)$. $Al(s) rarr Al^(3+) + 3e^(-)$ $(i)$ And cupric ion is reduced to zerovalent copper.... $Cu^(2+) + 2e^(-) rarr Cu(s)$ $(ii)$ We cross multiply to remove...

1 Answers 1 views
If 5.563 grams of $CH_4$ reacted with 25.00 grams of oxygen, how much $CO_2$ , in grams, can be produced? $CH_4(g) + 2O2(g) -> CO_2(g) + 2H_2O(g)$?

$"Moles of methane"$ $=$ $(5.563*g)/(16.043*g*mol^-1)$ $=$ $0.3468*mol$. $"Moles of dioxygen"$ $=$ $(25.00*g)/(31.998*g*mol^-1)$ $=$ $0.7813*mol$. Clearly there is sufficient dioxygen to react with methane as per the stoichiometric equation: $CH_4(g) +...

1 Answers 1 views
How do you determine the maximum number of moles of product that can be produced from 7.0 mol $Al$ and 8.0 mol $Cl_2$ according to the equation $2Al + 3Cl_2 -> 2AlCl_3$?

This is an example of a limiting reactant problem. The first step is to write down the balanced chemical equation. That is, $2Al(s)+3Cl_2(g)->2AlCl_3(s)$ We got seven moles of aluminium and...

1 Answers 1 views
How many grams of $AlCl_3$ can be produced by reacting 80.0 grams of $Cl_2$ with an unlimited supply of $Al$?

Your starting point here will be the balanced chemical equation for this . Aluminium metal, $"Al"$, will react with chlorine gas, $"Cl"_2$, to produce aluminium chloride, $"AlCl"_3$, according to the...

1 Answers 1 views
How many grams of aluminum are required to produce 8.70 moles of aluminum chloride in the reaction $2Al + 3Cl_2 -> 2AlCl_3$?

The balanced chemical equation represents ratio in which the chemicals combine, and in this case illustrates that $2 mol Al$ produces $2molAlCl_3$, hence these 2 chemicals are in a $1:1$...

1 Answers 1 views
$2Al + 3Cl_2 -> 2AlCl_3$ How many grams of chlorine are required to produce 4 moles of aluminum chloride?

$2Al(s) + 3Cl_2(g) rarr 2AlCl_3(s)$ You have given a stoichiometrically balanced equation, so bravo. The equation explicitly tells us that $54*g$ of aluminum metal reacts with $6xx35.45*g$ $Cl_2$ gas to...

1 Answers 1 views
Using the following reaction, if 5.93 grams of sodium cyanide are reacted with 8.45 grams of sulfuric acid, how many grams of sodium sulfate will be produced? $2NaCN + H_2SO_4 -> Na_2SO_4 + 2HCN$

This appears to be a limiting reactant problem. 1. Write the balanced equation $"2NaCN" + "H"_2"SO"_4 → "Na"_2"SO"_4 + "2HCN"$ 2. Calculate the moles of $"Na"_2"SO"_4$ formed from...

1 Answers 1 views
What volume of 133 mL, 7.90 M $CuCl_2$ solution should you dilute in order to have 51.5 mL of diluted solution that contains 4.49 g $CuCl_2$?

The first thing to do here is to use the and the volume of the initial solution to figure out how many grams of copper(II) chloride it contains....

1 Answers 1 views
X