Which formula can be used to calculate the theoretical yield? which formula can be used to calculate the theoretical yield?

step 1

determine the number of moles of each reactant. for solids, divide the mass of a reactant used by its molecular weight. for liquids and gases, multiply the volume by the density and then divide by the molecular weight.

step 2

multiply the molecular weight by the number of moles in the equation. the reactant that has the smallest mole number is the limiting reagent.

step 3

calculate the theoretical mole yield by using the chemical equation. the multiply the ratio between the limiting reagent and the product by the number of moles of the limiting reagent used in the experiment. for example, if your equation was 4al + 3o2 yields 2 al2o3, and al was your limiting reagent, you would divide the number of al moles used by two because it takes four moles of al to make two moles of al2o3, a ratio of two to one.

step 4

multiply the number of moles of the product by the molecular weight of the product to determine the theoretical yield. for example, if you created 0.5 moles of al2o3, the molecular weight of al2co3 is 101.96 g/mol, so you would get 50.98 grams as the theoretical yield.

brainliest is nice

why do the isotopes of a particular element behave differently in nuclear reactions but the same in chemical reactions? chemical reactions involve changes to electrons in an atom; isotopes do not differ in electrons, and so they react the same way during chemical changes.

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