100.0mL of 0.40M Al(NO3)3 is reacted with 100.0 mL of 0.40M Al(OH)3 according to
the following reaction: Al(NO3)3 + 3KOH → Al(OH)3 + 3KNO3

1) Identify the limiting reactant.
2) Identify the excess reactant.
3) Determine the theoretical yield of Al(OH)3
4) How much excess reactant is left over after the reaction is complete?
5) If 0.50g of Al(OH)3 determine the percent yield of Al(OH)3

Question:

100.0mL of 0.40M Al(NO₃)₃ is reacted with 100.0 mL of 0.40M KOH according to

the following reaction: Al(NO₃)₃ + 3KOH → Al(OH)₃ + 3KNO₃

1) KOH

2) Al(NO₃)₃

3) 0.01333 moles or 1.04004 g of Al(OH)₃

4) The amount of the excess reactant, Al(NO₃)₃ left after the reaction = 2/75 moles or 5.6799 grams of Al(NO₃)₃

5) The percentage yield of Al(OH)₃ is 48.075%

Explanation:

1) The reaction is given as follows;

Al(NO₃)₃ + 3KOH  → Al(OH)₃ + 3KNO₃

One mole of Al(NO₃)₃  reacts with three moles of 3KOH  to produce one mole of Al(OH)₃ and three moles of 3KNO₃

Hence when 0.4 moles of Al(NO₃)₃ reacts with 0.4 moles of KOH, the limiting reagent is found as follows;

1/3 moles Al(NO₃)₃  combines with 1 moles  of KOH

Hence 0.4/3 moles of Al(NO₃)₃  will combine with 0.4 moles  of KOH from which there will be an excess of (0.4 - 0.4/3) moles of Al(NO₃)₃ left, making KOH the limiting reactant

2) The excess reactant as shown above is the Al(NO₃)₃

3) From the reaction above, 0.4 moles of KOH combined with excess Al(NO₃)₃ will produce 0.4/3 moles of Al(OH)₃ however, since the reagent are each 100 mL, and 0.40 M indicates 0.4 moles per 1000 mL the number of moles of Al(OH)₃ produced is given as follows;

0.4/3 = 2/15 moles per 1000 mL ≡ 2/15 ÷ 10 = 2/150 moles per 1000 mL ÷ 10 or 100 mL

The number of moles of Al(OH)₃ produced = 2/150 = 0.01333 moles

Molar mass of Al(OH)₃ = 78.003 g/mol

Mass of Al(OH)₃ produced = Number of moles produced × Molar mass

Mass of Al(OH)₃ produced = 0.01333 × 78.003 = 1.04004 g

4) The amount of the excess reactant, Al(NO₃)₃ left after the reaction is found as follows;

0.4 - 0.4/3 = 2/5 - 2/15 = 4/15 moles of Al(NO₃)₃ per 1000 mL or 4/(15×10) =2/75 moles per 100 mL

Molar mass of  Al(NO₃)₃ 212.996238 g/mol

Mass left = Number of moles × Molar mass = 5.6799 grams

The amount of the excess reactant, Al(NO₃)₃ left after the reaction = 2/75 moles or 5.6799 grams of Al(NO₃)₃

5) If 0.50 g of  Al(OH)₃ is produced, we have

Actual yield = 0.01333 moles of Al(OH)₃

Molar mass of Al(OH)₃ = 78.003 g/mol

Mass of theoretical yield = Molar Mass × Theoretical yield =  0.01333 × 78.003 = 1.04004 g of Al(OH)₃

Therefore;

The percentage yield of Al(OH)₃ = 48.075%.

most historians agree that gabriel mouton, the vicar of st. paul’s church in lyons, france, is the “founding father” of the metric system. he proposed a decimal system of measurement in 1670. mouton based it on the length of one minute of arc of a great circle of the earth (now called a nautical mile, 1852 meters). he also proposed the swing-length of a pendulum with a frequency of one beat per second as the unit of length (about 25 cm). a pendulum beating with this length would have been fairly easy to produce, thus facilitating the widespread distribution of uniform standards. over the years, his work was revised, improved, and extended by a number of french scientists.

the political sponsor of weights and measures reform in the french revolutionary national assembly was the bishop of autun, better known as talleyrand. under his auspices, the french academy appointed several committees to carry out the work of developing a usable system of weights and measures for france. one of the committees recommended a decimalized measurement system based upon a length equal to one ten-millionth of the length of a quadrant of the earth’s meridian (i.e., one ten-millionth of the distance between the equator and the north pole).

it is b that is the answer.