What mass of NaCl formed when 0.25 g Na react completely with $0.39 g Cl _2$?
A. -0.14 g NaCl
B. 0.14 g NaCl
C. 0.59 g NaCl
D. 0.64 g NaCl
Answer (1)
Determine the limiting reactant by calculating the moles of Na and C l 2 and comparing their mole ratio to the balanced chemical equation.
Calculate the moles of Na: n N a = 22.99 0.25 = 0.01087 mol and the moles of C l 2 : n C l 2 = 2 × 35.45 0.39 = 0.00550 mol.
Since Na is the limiting reactant, the moles of NaCl formed equals the moles of Na: n N a Cl = 0.01087 mol.
Calculate the mass of NaCl formed: ma s s N a Cl = 0.01087 × ( 22.99 + 35.45 ) = 0.635 g, so the answer is 0.64 g N a Cl .
Explanation
Balanced Chemical Equation First, we need to determine the limiting reactant in the reaction between Na and C l 2 to form NaCl. The balanced chemical equation is:
2 N a + C l 2 → 2 N a Cl
This tells us that 2 moles of Na react with 1 mole of C l 2 to produce 2 moles of NaCl.
Calculating Moles Next, we calculate the number of moles of Na and C l 2 using their respective molar masses:
Molar mass of Na (sodium) = 22.99 g/mol Molar mass of C l 2 (chlorine gas) = 2 * 35.45 = 70.90 g/mol
Moles of Na = M N a ma s s N a = 22.99 g / m o l 0.25 g = 0.01087 mol Moles of C l 2 = M C l 2 ma s s C l 2 = 70.90 g / m o l 0.39 g = 0.00550 mol
Determining Limiting Reactant Now, we determine the limiting reactant. According to the balanced equation, 2 moles of Na react with 1 mole of C l 2 . Therefore, the mole ratio of Na to C l 2 is 2:1.
We have 0.01087 mol of Na and 0.00550 mol of C l 2 . To determine which is the limiting reactant, we can calculate how much C l 2 is required to react completely with the available Na:
Required moles of C l 2 = 2 m o l e s N a = 2 0.01087 m o l = 0.005435 mol
Since we have 0.00550 mol of C l 2 , which is more than the 0.005435 mol required to react with all the Na, Na is the limiting reactant.
Moles of NaCl Formed Since Na is the limiting reactant, the number of moles of NaCl formed is equal to the number of moles of Na reacted:
Moles of NaCl = Moles of Na = 0.01087 mol
Calculating Mass of NaCl Finally, we calculate the mass of NaCl formed using its molar mass:
Molar mass of NaCl = 22.99 g/mol (Na) + 35.45 g/mol (Cl) = 58.44 g/mol
Mass of NaCl = Moles of NaCl * Molar mass of NaCl = 0.01087 mol * 58.44 g/mol = 0.635 g
Final Answer Therefore, the mass of NaCl formed when 0.25 g of Na reacts completely with 0.39 g of C l 2 is approximately 0.635 g. The closest answer from the given options is 0.64 g NaCl.
Examples
In the manufacturing of table salt (NaCl), it's crucial to know the exact quantities of sodium (Na) and chlorine ( C l 2 ) needed to produce a specific amount of salt. This calculation helps ensure that the reaction is efficient and that minimal resources are wasted. For instance, if a company wants to produce 100 kg of NaCl, they need to calculate the precise amounts of Na and C l 2 required, considering factors like purity and yield. Accurate stoichiometric calculations are essential for cost-effective and sustainable production in the chemical industry. This principle extends to various chemical processes, from pharmaceuticals to plastics, where precise control over reactant quantities is vital for product quality and safety.
