What is the percentage of insoluble sample in the unknown if the original sample had a mass of 2.732 g and that of the insoluble solid was determined to be 1.325 g? don't include the % sign in your answer.

The percentage of insoluble solid in the unknown is 48.5

Explanation:

To calculate the percentage of insoluble solid in the unknown sample, we use the equation:

Mass of unknown sample = 2.732 g

Mass of insoluble solid = 1.325 g

Putting values in above equation, we get:

Hence, the percentage of insoluble solid in the unknown is 48.5

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

the answer is a and have a great day