Question

What is the experimental yield (in g of precipitate) when 16.6 mL of a 0.5 M solution of sodium chloride is combined with 17.2 mL of a 0.567 M solution of silver nitrate at a 89.8% yield?

What is the theoretical yield (in g of precipitate) when 17.6 mL of a 0.501 M solution of iron(III) chloride is combined with 17.9 mL of a 0.614 M solution of lead(II) nitrate?

Nitrogen and oxygen react to form nitric oxide (NO). All substances are in the gas phase. If 0.448 atm of nitrogen and 0.371 atm of oxygen react, what is the partial pressure of nitric oxide (in mmHg) when this reaction goes 74.5 complete. The temperature and volume are constant.

Answer 1

1) The balanced chemical equation for the reaction is

NaCl (aq) + AgNO₃ (aq) -------> AgCl (s) + NaNO₃ (aq)

As per the stoichiometry of the reaction,

1 mole NaCl = 1 mole AgNO₃.

Determine the millimoles of NaCl and AgNO₃ used in the reaction.

Millimoles of NaCl = (16.6 mL)*(0.5 M) = 8.3 mmole.

Millimoles of AgNO₃ = (17.2 mL)*(0.567 M) = 9.7524 mmole.

Offcourse, NaCl is the limiting reactant and the yield of the product is decided by the mole(s) of NaCl used.

As per the stoichiometric equation,

1 mole NaCl = 1 mole AgCl.

Therefore, 8.3 mmole NaCl = 8.3 mmole AgCl. This is obtained considering 100% yield, i.e, 100% conversion of the limiting reactant into the desired product. However, the yield is capped at 89.9%; therefore, the actual yield of AgCl = (89.8%)*(8.3 mmole) = (89.8/100)*(8.3 mmole) = 7.4534 mmole.

Molar mass of AgCl = (1*107.8682 + 1*35.453) g/mol = 143.3212 g/mol.

Mass of AgCl obtained = (7.4534 mmole)*(1 mole/1000 mmole)*(143.3212 g/mol) = 1.0682 g (ans).