Lead(II) sulfide was once used in glazing earthenware. It will also react with hydrogen peroxide to form lead(II) sulfate and water. How many grams of hydrogen peroxide are needed to react completely with 265 g of lead(II) sulfide?

The mass of hydrogen peroxide needed to react completely with given amount of lead (II) sulfide is 122.9 grams.

Explanation:

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

    .....(1)

Given mass of lead(II) sulfide = 265 g

Molar mass of lead(II) sulfide = 293.3 g/mol

Putting values in equation 1, we get:

The chemical equation for the reaction of lead sulfide and hydrogen peroxide follows:

By Stoichiometry of the reaction:

1 mole of lead(II) sulfide reacts with 4 moles of hydrogen peroxide

So, 0.904 moles of lead(II) sulfide will react with = of hydrogen peroxide

Now, calculating the mass of hydrogen peroxide from equation 1, we get:

Molar mass of hydrogen peroxide = 34 g/mol

Moles of hydrogen peroxide = 3.616 moles

Putting values in equation 1, we get:

Hence, the mass of hydrogen peroxide needed to react completely with given amount of lead (II) sulfide is 122.9 grams.

a boiling water reactor does not boil the water. a pressurized water reactor boils the water faster. a pressurized water reactor is heated directly by the core. a pressurized water reactor does not use fission.

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