Question

What is the % yield if 17.4g of sodium hydroxide is produced when 20.0g of sodium metal reacts with 19g of water according to the chemical equation 2Na(s) + 2H₂O(l) → 2NaOH(aq)+ H₂(g)?

Answer 1

50.05% is the percent yield if 17.4g of sodium hydroxide is produced when 20.0g of sodium metal reacts with 19g of water.

Explanation:

Balanced chemical equation for the reaction:

2Na + 2 $H_{2}$O ⇒ 2NaOH + $H_{2}$

Data given:

mass of NaOH produced = 17.4 grams

mass of Na reacted = 20 grams

mass of water reacted = 19 grams

percent yield =?

atomic mass of Na = 23 grams/mole

atomic mass of water = 18 grams/mole

atomic mass of NaOH = 39.9 grams/mole

number of moles is calculated as:

number of moles = $\frac{mass}{atomic mass of 1 mole}$

putting the values in above equation to know number of moles:

number of moles of Na = $\frac{20}{23}$

                                       = 0.869 moles

number of moles of water = $\frac{19}{18}$

                                            = 1.05 mole

number of moles of NaOH = $\frac{17.4}{40}$

                                             = 0.435 moles

limiting reagent in the reaction is Na

2 moles of Na reacted to give 2 moles of NaOH

So, 0.869 moles of NaOH will give 0.869 moles of NaOH

Grams of NaOH = 34.76 grams (theoretical yield)

2 moles of water will give 2 moles of NaOH

Hence, 1.05 moles will give 1.05 moles of NaOH

grams of NaOH = 42 grams

percent yield = $\frac{actual yield}{theoretical yield}$

putting the values in the equation:

percent yield = $\frac{17.4}{34.76}$ x100

                     = 50.05 %

50.05 % is the percent yield.