Question

REACTANTS 25 grams of sodium hydroxide combines with 18 grams of hydrochloric acid to form 26.5 grams of salt and water. Please answer the following questions? A) What is the limiting reactant? B) How many grams of sodium chloride are produced? C) What is the percent yield of the sodium chloride? D) How much reactant is used? E) How much reactant is in excess?

Answer 1

Let us write the balanced equation:
HCl + NaOH =====> NaCl + H₂O
Reaction type: double replacement

Mass of NaOH = 25 g

Moles of NaOH = 25 g /40 g/mol =  0.625 Moles

Mass of HCl = 18 g

Moles of HCl = 18 g /36.45 g/mol = 0.4936 Moles

Limiting reagent is HCl which consumed completely

Moles of Salt produced = 0.4936 Moles

Theoretical Mass of salt produced = 0.4936 mole x 58.45 g/mol =  28.85 g

Actual mass of salt obtained = 26.5 g

Percentage yield = 26.5 g x 100 % / 28.85 g = 91.85 %

Excess reagent is NaOH  

Mass of NaOH left after reaction = 25 g - 19.746 g = 5.254 g

Using the above calculation we can answer the given question s

Question A Limiting reagent is HCl

Question B 28.85 g of NaCl is produced

Question C percentage yield is 91.85 %

Question d HCl used compeletly and 19.746 g of NaOH used

Question E 5.254 g of NaOH will be excess after reaction completion

Balanced equation: HCI + NaOH = NaCl + H2O Reaction type: double replacement Reaction stoichiometry Limiting reagent Compound Coefficient Molar Mass Moles Weight HCI 36.46094 0.4936789 18.00000 NaOH 39.99710928 0.4936789 19.74573 NaCl 58.44276928 0.4936789 28.85196 H2O 18.01528 0.4936789 8.893764 Units: molar mass - g/mol, weight...