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%

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.