For a theoretical yield of 2.97 g and percent yield of 93.4452%, calculate the actual yield for a chemical reaction. Answer in units of g.

2.77532244 grams

Explanation:

Percent yeild=actual/theoretical * 100

Plug in the values

93.4452=actual/2.97 *100

0.934452*2.97=2.77532244

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 electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.