What mass of NaCl formed when 0.25 g Na react completely with $0.39 g Cl _2$?
A. -0.14 g NaCl
B. 0.64 g NaCl
C. 0.14 g NaCl
D. 0.59 g NaCl
Answer (1)
Calculate the moles of Na and C l 2 using their respective masses and molar masses.
Determine the limiting reactant by comparing the mole ratio of Na and C l 2 to the stoichiometric ratio in the balanced equation.
Since Na is the limiting reactant, the moles of NaCl formed equals the moles of Na.
Calculate the mass of NaCl formed using the moles of NaCl and its molar mass, resulting in approximately 0.64 g.
Explanation
Balanced Chemical Equation First, we need to identify 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
Calculate Moles of Reactants Next, we calculate the number of moles of Na and C l 2 :
n N a = M N a ma s s N a = 22.99 g / m o l 0.25 g = 0.01087 m o l n C l 2 = M C l 2 ma s s C l 2 = 70.90 g / m o l 0.39 g = 0.00550 m o l
Identify 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, we need to compare the mole ratio: n C l 2 n N a = 0.00550 0.01087 = 1.976 Since this ratio is less than 2, 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: n N a Cl = n N a = 0.01087 m o l
Calculate Mass of NaCl Finally, we calculate the mass of NaCl formed: ma s s N a Cl = n N a Cl ∗ M N a Cl = 0.01087 m o l ∗ 58.44 g / m o l = 0.635 g
Final Answer Therefore, the mass of NaCl formed is approximately 0.64 g.
Examples
In real-world applications, stoichiometry is crucial in chemical manufacturing. For instance, in producing pharmaceuticals, ensuring the correct mass of reactants like sodium and chlorine is vital to synthesize the desired amount of sodium chloride, a key ingredient in saline solutions. Accurate calculations prevent waste and ensure product quality. This principle extends to various industries, including food production, where precise salt concentrations are essential for preservation and taste.
