Final answer:
c. 188 g. From the reaction Cl₂ + 2 KBr → 2 KCl + Br₂, using stoichiometry and the concept of limiting reagent, we find that we can produce 188 g of potassium chloride from 300 g of chlorine and 300 g of potassium bromide.
Step-by-step explanation:
To determine how many grams of potassium chloride can be produced from 300 g of chlorine and 300 g of potassium bromide, we must first write out the balanced chemical equation:
Cl2 + 2 KBr → 2 KCl + Br2
According to the stoichiometry of the reaction, 1 mole of Cl2 reacts with 2 moles of KBr to produce 2 moles of KCl. To solve this problem, we need to find the limiting reagent, which is the reactant that will be completely consumed first and thus will limit the amount of products formed.
Using the molar masses (Cl2 = 70.906 g/mol, KBr = 119.002 g/mol, KCl = 74.551 g/mol), we can convert the given mass of each reactant to moles:
- 300 g Cl2 / 70.906 g/mol = 4.23 moles Cl2
- 300 g KBr / 119.002 g/mol = 2.52 moles KBr
Comparing the mole ratio, we see that KBr is the limiting reagent. So, 2.52 moles of KBr will produce 2.52 moles of KCl. Now we convert moles of KCl to grams:
- 2.52 moles KCl * 74.551 g/mol = 187.93 g KCl
Hence, we can produce 187.93 g of potassium chloride, which can be rounded to 188 g to match one of the answer options.