4. Cobalt(II) chloride reacts with fluorine in a single replacement reaction to produce cobalt(II) fluoride and chlorine gas.

Balanced Equation:

+

How many moles of fluorine are required to produce 290.8 g of cobalt(II) fluoride?

tinn

3.000 moles of flourine are required to produce 290.8 g of cobalt(II) fluoride

Explanation:

A single replacement reaction is one in which a more reactive element displaces a less reactive element from its salt solution.  

When cobalt chloride is added to flourine, chlorine being more reactive than flourine, displaces chlorine atom its salt solution and lead to formation of cobalt flouride and chlorine gas.

To calculate the number of moles, we use the equation:

According to stoichiometry :

1 mole of require = 1 mole of

Thus 3.000 moles of will require=  of

 Thus 3.000 moles of flourine are required to produce 290.8 g of cobalt(II) fluoride

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

the answer is 75.0%.