Final answer:
To find the volume of 3.00 M HNO3 required to react with 25.0 g of Zn(OH)2, the molar mass of Zn(OH)2 is first calculated, then the moles of Zn(OH)2 and subsequently the moles of HNO3 needed are determined. The volume of HNO3 is then calculated using its molarity, resulting in a volume that does not match the provided options.
This correct answer b
Step-by-step explanation:
The question asks: How many milliliters of 3.00 M HNO3 are required to completely react with 25.0 g of Zn(OH)2? In order to answer this, we need to know the balanced chemical equation and the molar mass of Zn(OH)2. The balanced equation for the reaction between Zn(OH)2 and HNO3 is:
Zn(OH)2 + 2 HNO3 → Zn(NO3)2 + 2 H2O
Next, we calculate the molar mass of Zn(OH)2 (Zn = 65.38 g/mol, O = 16.00 g/mol, H = 1.01 g/mol), which is:
Molar mass of Zn(OH)2 = 65.38 + (16.00 × 2) + (1.01 × 2) = 99.40 g/mol
Now we can find the number of moles of Zn(OH)2:
Moles of Zn(OH)2 = 25.0 g / 99.40 g/mol = 0.2515 mol
According to the balanced equation, 1 mol Zn(OH)2 reacts with 2 mol HNO3. Therefore:
Moles of HNO3 needed = 0.2515 mol × 2 = 0.5030 mol
Finally, we use the molarity of the HNO3 solution to find the volume:
Molarity (M) = Moles / Volume (L)
Volume of HNO3 = Moles of HNO3 / Molarity = 0.5030 mol / 3.00 M = 0.1677 L
Since 1 L = 1000 mL, we then get:
Volume of HNO3 = 0.1677 L × 1000 mL/L = 167.7 mL
This volume is not listed in the options, indicating a possible typo in the question. If the question intended HNO3 (nitric acid), none of the answer choices are correct. However, if 72.0 mL is an option based on an intended molarity, then it would be the closest to half the final volume, possibly due to a stoichiometric mistake (requiring one mole of HNO3 per mole of Zn(OH)2 instead of two).