Aluminum bromide reacts with chlorine gas to produce aluminum chloride and liquid bromide,
according to the chemical equation below:
2aibr+ 3c1, 2aici, + 3br2
how many moles of aluminum chloride are made if 779 g of chlorine gas are consumed?

7.5 mol of AlCl₃ will produce.

Explanation:

Given data:

Mass of chlorine = 779 g

Moles of aluminium chloride = ?

Solution:

Chemical equation:

2AlBr₃ + 3Cl₂ → 2AlCl₃ + 3Br₂

Number of moles of Cl₂:

Number of moles = mass/ molar mass

Number of moles = 779 g / 71 g/mol

Number of moles = 11.3 mol

Now we will compare the moles of chlorine with AlCl₃ .

                        Cl₂       :       AlCl₃

                         3         :        2

                        11.3      :      2/3×11.3 = 7.5 mol

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

i'm pretty sure its b i might be wrong though!