Can someone please help me work out the theoretical moles and the actual moles,

Question

Can someone please help me work out the theoretical moles and the actual moles, along with the percent yeild given the following information

1 15 points Data table is complete and readable, and values have correct sig figs. -1 point One point each is deducted for incorrect sig figs, messiness, and every blank tat cell. Table 1. Data and Results table. Initial mass of vial containing CusO, 5H2O (2) Final mass of vial containing CuSO4 5H20 (g,) Mass of CuSO4 5H20 (g) Moles CusO4 5H2O (mol) Volume of NH3 added (mL) Moles of of NH3 (mol) Theoretical moles of [Cu(NH3)4] SO4 H2O (mol) Mass of product beaker (g) Mass of beaker and product (g) Mass of product (g) Actual moles of [Cu(NH3)4] SO4 H2O (mol) Percent yield 1.471 al.o mL Lillmoles 12,3 73

Explanation / Answer

The reaction is given by

CuS04 .5H20   + 4Nh3   ----> [ Cu(NH3) 4 ] S04 .H20   + 4H20

from the reaction

1 mole of CuS04 .5H20 reacts with 4 moles of NH3

let   0.005972 moles of Cus04.5H20 react with y moles of NH3

y = 4 x 0.005972 = 0.02388

so 0.02388 moles of Nh3 is required .

so excess Nh3 is in excess .

from the reaction

1 mole of Cus04.5h20 on reaction gives 1 mole of [ Cu(NH3) 4 ] S04 .H20

so 0.005972 mole of Cus04.5h20 on reaction gives 0.005972 mole of [ Cu(NH3) 4 ] S04 .H20

theoretical yield of   [ Cu(NH3) 4 ] S04 .H20 = 0.005972 moles

molar mass of [ Cu(NH3) 4 ] S04 .H20 = 245.79

given mass of [ Cu(NH3) 4 ] S04 .H20 = 1.4081 g

moles = mass / molar mass

moles = 1.4081 / 245.79

moles of [ Cu(NH3) 4 ] S04 .H20 = 0.0057288

   so   actual yield of [ Cu(NH3) 4 ] S04 .H20 = 0.0057288 moles

percent yield = actual / theoretical   x 100

percent yield = 0.0057288 / 0.005972    x 100

percent yield = 95.93

so percent yield is 95.93 %

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---