A student synthesized a cobalt amine complex with 11.0323 g of CoCl2·6H2O (limiting reactant, MW = 237.83 g/mol) and collected 9.4705 g of the cobalt amine complex (MW = 250.45 g/mol). What was the percent yield?

The percent yield would be 81.51%

Explanation:

Number of moles of CoCl2·6H2O would be calculated by using the formula

number of moles = given weight/molecular weight.

Thus, by putting values,

= 11.0323/237.83

= 0.046387

Number of moles of CO+2 = Number of moles of CoCl2·6H2O = 0.046387

As we know that CoCl2·6H2O is a limiting reactant, therefore, the number of moles of complex = Number of moles of CO+2 = 0.046387.

Hence, we can calculate the mass of the complex as follows:

Molecular weigth x number of moles = 250.45 x 0.046387

Thus, the theoratical yield would be 11.6176.

We know that actual yield as given in statement is 9.4705 g. Therefore, percent yield would be,

= (9.4705 / 11.6176) x 100

= 81.51%

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