All categories

3 years ago

What is the limiting reactant if 0.5 g Al is reacted with 3.5 g CuCl2?

1 answer:

6 0

The balanced equation for the above reaction is
2Al + 3CuCl₂ --> 2AlCl₃ + 3Cu
stoichiometry of Al to CuCl₂ is 2:3
limiting reactant is when the reactant is fully consumed in the reaction therefore amount of product formed depends on amount of limiting reactant present.
number of Al moles - 0.5 g / 27 g/mol = 0.019 mol
number of CuCl₂ moles - 3.5 g / 134.5 g/mol = 0.026 mol
if Al is the limiting reactant
if 2 mol of Al reacts with 3 mol of CuCl₂
then 0.019 mol of Al reacts with - 3/2 x 0.019 = 0.029 mol of CuCl₂
but only 0.026 mol of CuCl₂ is present
therefore CuCl₂ is the limiting reactant
and 0.026 mol of CuCl₂ reacts with - 0.026/3 x 2 = 0.017 mol of Al is required
but 0.019 mol of Al is present
therefore CuCl₂ is the limiting reactant and Al is in excess

You might be interested in

What was the average speed of the truck for this trip?

Goshia [24]

Answer:

31.67 mph

Explanation:

To calculate the average speed of the truck, we must first obtain the total distance travelled by the truck followed by the total time taken for the truck to cover the distance travelled.

The following data were obtained from the question include:

Total distance) = 30 + 45 + 50 + 65 = 190 miles

Total time = 1 + 2 +1 +2 = 6 hours

Average speed =.?

Average speed = Total distance / Total time

Average speed = 190 /6

Average speed = 31.67 mph

Therefore, the average speed of the truck is 31.67 mph

8 0

4 years ago

How many moles are in 35.5 g of aluminum?

Blizzard [7]

Answer:

1.316 moles of Aluminum

Explanation:

The molar mass for Al is 26.98 g/mol

$\frac{35.5g}{26.98g/mol} =1.316 moles Al$

3 0

3 years ago

Calculate the molarity of a solution obtained dissolving 10.0 g of cobalt(Ⅱ) bromide tetrahydrate in enough water to make 450 mL

Vladimir [108]

Answer:

The molarity of the CoBr2•4H2O solution is 7.64 × 10-2 M

Explanation:

Cobalt (II) bromide tetrahydrate

• Cobalt - A transition metal with Roman numeral (II) → charge: +2 → Co2+

• Bromide - anion from group 7A → -1 charge → symbol: Br-

• Tetrahydrate- tetra- means 4 and hydrate is H2O

The chemical formula of the compound is: CoBr2•4H2O

We then need to determine the number of moles of CoBr2•4H2O since this is the only information missing for us to find molarity. Notice that the volume of the solution is already given.

We’re given the mass of CoBr2•4H2O. We can use the molar mass of CoBr2•4H2O4 to find the moles.

•The molar mass of CoBr2•4H2O is:

CoBr2•4H2O

1 Co x 58.93 g/mol Co = 58.93 g/mol

2 Br x 79.90 g/mol Br = 159.80 g/mol

8 H x 1.008 g/mol H = 8.064 g/mol

4 O x 16.00 g/mol O = 64.00 g/mol

________________________________________

Sum = 290.79 g/ mo

The moles of CoBr2•4H2O is:

= 10.0 g CoBr2•4H2O x $\frac{ 1 mol CoBr_2 . 4H_2O}{290.79 g CoBr_2 . 4H_2O}$

= 0.0344 mol CoBr2•4H

We know that the volume of the solution is 450 mL.

We can now calculate for molarity:

Convert mL to L → 1 mL = 10-3 L

Formula:

Molarity (M)= Mole of solute / Liters of solution

= 0.0344 mol CoBr2•4H / 450 mL x 1 ml / 10^ -3 L

= 0.0764

= 7.64 × 10-2 mol/L

8 0

4 years ago

A reaction A(aq)+B(aq)↽−−⇀C(aq) has a standard free‑energy change of −4.20 kJ/mol at 25 °C. What are the concentrations of A, B,

Dafna1 [17]

Answer : The concentration of $A,B\text{ and }C$ at equilibrium are 0.132 M, 0.232 M and 0.168 M respectively.

Explanation :

The given chemical reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

First we have to calculate the equilibrium constant for the reaction.

The relation between the equilibrium constant and standard free‑energy is:

$\Delta G^o=-RT \ln k$

where,

$\Delta G^o$ = standard free‑energy change = -4.20 kJ/mole

R = universal gas constant = 8.314 J/mole.K

k = equilibrium constant = ?

T = temperature = $25^oC=273+25=298K$

Now put all the given values in the above relation, we get:

$-4.20kJ/mole=-(8.314J/mole.K)\times (298K) \ln k$

$k=5.45$

Now we have to calculate the concentrations of A, B, and C at equilibrium.

The given equilibrium reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

Initially 0.30 0.40 0

At equilibrium (0.30-x) (0.40-x) x

The expression of equilibrium constant will be,

$k=\frac{[C]}{[A][B]}$

$5.45=\frac{x}{(0.30-x)\times (0.40-x)}$

By solving the term x, we get

$x=0.168\text{ and }0.716$

From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.

The value of x will be, 0.168 M

The concentration of $A$ at equilibrium = (0.30-x) = 0.30 - 0.168 = 0.132 M