A laboratory technician combined sodium hydroxide with excess iron(II) nitrate. A reaction took place according to this chemical equation: $2 NaOH + Fe ( NO _3)_2 \rightarrow NaNO _3+ Fe ( OH )_2$. The reaction produced 3.70 grams of iron(II) hydroxide. Assuming the reaction came to completion, what was the initial mass of sodium hydroxide?
A. 1.6 g
B. 2.0 g
C. 3.3 g
Answer (2)
Calculate the molar mass of F e ( O H ) 2 : M F e ( O H ) 2 = 89.859 g/mol.
Calculate the moles of F e ( O H ) 2 : m o l e s F e ( O H ) 2 = 0.0411756 mol.
Determine the moles of N a O H : m o l e s N a O H = 2 × 0.0411756 = 0.0823512 mol.
Calculate the mass of N a O H : ma s s N a O H = 0.0823512 × 39.997 = 3.2938 g. The answer is 3.3 g .
Explanation
Problem Analysis We are given the balanced chemical equation:
2 N a O H + F e ( N O 3 ) 2 → 2 N a N O 3 + F e ( O H ) 2
We know that 3.70 grams of F e ( O H ) 2 were produced, and we need to find the initial mass of N a O H used in the reaction.
Molar Mass of Iron(II) Hydroxide First, we need to calculate the molar mass of F e ( O H ) 2 . Using the periodic table, we have:
Iron (Fe): 55.845 g/mol
Oxygen (O): 15.999 g/mol
Hydrogen (H): 1.008 g/mol
So, the molar mass of F e ( O H ) 2 is:
M F e ( O H ) 2 = 55.845 + 2 ( 15.999 + 1.008 ) = 55.845 + 2 ( 17.007 ) = 55.845 + 34.014 = 89.859 g/mol
Moles of Iron(II) Hydroxide Next, we calculate the number of moles of F e ( O H ) 2 produced:
m o l es = m o l a r _ ma ss ma ss
m o l e s F e ( O H ) 2 = 89.859 g/mol 3.70 g = 0.0411756 mol
Moles of Sodium Hydroxide From the balanced chemical equation, the mole ratio between N a O H and F e ( O H ) 2 is 2:1. This means that for every 1 mole of F e ( O H ) 2 produced, 2 moles of N a O H were required.
Therefore, the number of moles of N a O H required is:
m o l e s N a O H = 2 × m o l e s F e ( O H ) 2 = 2 × 0.0411756 mol = 0.0823512 mol
Molar Mass of Sodium Hydroxide Now, we calculate the molar mass of N a O H :
Sodium (Na): 22.990 g/mol
Oxygen (O): 15.999 g/mol
Hydrogen (H): 1.008 g/mol
So, the molar mass of N a O H is:
M N a O H = 22.990 + 15.999 + 1.008 = 39.997 g/mol
Mass of Sodium Hydroxide Finally, we calculate the mass of N a O H required:
ma ss = m o l es × m o l a r _ ma ss
ma s s N a O H = 0.0823512 mol × 39.997 g/mol = 3.2938 g
Final Answer Therefore, the initial mass of sodium hydroxide is approximately 3.29 g. The closest answer choice is C. 3.3 g.
Examples
In chemistry, stoichiometry is essential for calculating the amounts of reactants needed or products formed in a chemical reaction. For example, in the production of pharmaceuticals, it's crucial to determine the precise amount of each ingredient to ensure the drug's efficacy and safety. Similarly, in environmental science, stoichiometry helps in calculating the amount of chemicals needed to neutralize pollutants in water or air, ensuring effective and safe remediation processes. By understanding mole ratios and molar masses, we can accurately predict and control chemical reactions in various real-world applications.
The initial mass of sodium hydroxide used in the reaction is approximately 3.3 grams. This is calculated from the amount of iron(II) hydroxide produced and the stoichiometry of the reaction. Hence, the correct answer is C. 3.3 g.
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