What mass of solid aluminum hydroxide is produced when 50.0 mL of 0.20 M aluminum nitrate is added to 200.0 mL of 0.100 M potassium hydroxide?

0.52g of Al(OH)₃

Explanation:

The reaction of aluminium nitrate, Al(NO₃)₃, reacts with KOH, as follows:

Al(NO₃)₃ + 3KOH → Al(OH)₃ + 3 KNO₃

That means 1 mole of aluminium nitrate reacts with 3 moles of potassium hydroxide.

To find moles of aluminium hydroxide we need to determine the limiting reactant finding moles of each reactant:

Moles aluminium nitrate:

0.050L * (0.20mol / L) = 0.01 moles Al(NO₃)₃

Moles KOH:

0.200L * (0.100mol / L) = 0.02 moles of KOH

Thus, for a complete reaction of 0.02 moles of KOH are needed (Using the chemical equation)

0.02 moles of KOH * (1 mol Al(NO₃)₃ / 3 mol KOH) = 0.0067 moles of Al(NO₃)₃.

As there are 0.01 moles of Al(NO₃)₃, this is the excess reactant and KOH is the limiting reactant.

The 0.02 moles of KOH produce:

0.02 moles of KOH * (1 mol Al(OH)₃ / 3 mol KOH) = 0.0067 moles of Al(OH)₃

In mass (Using molar mass of Al(OH)₃: 78g/mol):

0.0067 moles of Al(OH)₃ * (78g/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

the answer is 1 i believe