Now examine the reaction between iron(III) chloride and sodium hydroxide. Use 60 g NaOH and add FeCl3. What mass in grams of FeCl3 must be added to just consume 60 g NaOH?

The mass of iron (III) chloride that must be added is 81.1 grams

Explanation:

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

    .....(1)

Given mass of NaOH = 60 g

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

The chemical equation for the reaction of iron (III) chloride and NaOH follows:

By Stoichiometry of the reaction:

3 moles of NaOH reacts with 1 mole of iron (III) chloride

So, 1.5 moles of NaOH will react with = of iron (III) chloride

Now, calculating the mass of iron (III) chloride from equation 1, we get:

Molar mass of iron (III) chloride = 162.2 g/mol

Moles of iron (III) chloride = 0.5 moles

Putting values in equation 1, we get:

Hence, the mass of iron (III) chloride that must be added is 81.1 grams

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.