Find the mass of sodium hydroxide needed to neutralize enough nitric acid to create 2.0 grams of sodium nitrate? How do I write out equation and solve?
See your post above.
To find the mass of sodium hydroxide (NaOH) needed to neutralize nitric acid (HNO3) and create a specific amount of sodium nitrate (NaNO3), you need to follow a few steps:
1. Write out the balanced chemical equation for the reaction between NaOH and HNO3:
NaOH + HNO3 -> NaNO3 + H2O
2. Determine the molar ratio between NaOH and HNO3 by looking at the coefficients in the balanced equation. In this case, the molar ratio is 1:1, meaning that one mole of NaOH reacts with one mole of HNO3.
3. Convert the given mass of sodium nitrate (NaNO3) to moles. To do this, divide the given mass by the molar mass of NaNO3. The molar mass of NaNO3 is the sum of the atomic masses of sodium (Na), nitrogen (N), and oxygen (O) in NaNO3.
4. Since the molar ratio between NaOH and HNO3 is 1:1, the number of moles of NaOH needed to neutralize the same number of moles of HNO3 is the same. Therefore, the number of moles of NaOH required is the same as the number of moles of NaNO3 calculated in step 3.
5. Finally, convert the calculated moles of NaOH to mass by multiplying it by the molar mass of NaOH. The molar mass of NaOH is the sum of the atomic masses of sodium (Na), oxygen (O), and hydrogen (H) in NaOH.
Let's perform the calculations:
1. Writing out the balanced equation: NaOH + HNO3 -> NaNO3 + H2O
2. Molar ratio: 1:1 (NaOH:HNO3)
3. Given mass of sodium nitrate (NaNO3): 2.0 grams
Molar mass of NaNO3: 22.99 g/mol (Na) + 14.01 g/mol (N) + 48.00 g/mol (O) = 85.00 g/mol
Moles of NaNO3 = Given mass / Molar mass
= 2.0 g / 85.00 g/mol = 0.0235 mol
4. Moles of NaOH needed = Moles of NaNO3 = 0.0235 mol
5. Molar mass of NaOH: 22.99 g/mol (Na) + 16.00 g/mol (O) + 1.01 g/mol (H) = 40.00 g/mol
Mass of NaOH = Moles of NaOH * Molar mass
= 0.0235 mol * 40.00 g/mol = 0.94 grams
Therefore, the mass of sodium hydroxide needed to neutralize enough nitric acid to create 2.0 grams of sodium nitrate is 0.94 grams.