The following questions are about the reaction of sodium oxide with nitric acid.
1)What mass of sodium oxide would be needed to react with acid to form 0.5 mol salt?
I said 1.5
2) How much sodium oxide would be required to neutralise 230cm3 of 1.0moldm-3?
3) 2.0gof sodium oxide was DDED TO 230CM3 OF 0.05MOLDM-3 HNO3 solution. Once all the acid has reacted the remaining sodium oxide will react with water to form an alkali.
a)What mass of alkali will be formed?
b)What will the concentration of the alkali solution be?
Please help i'm really confused and tell me please if i got 1 right
is the last one 0.0416
Na2O + 2HNO3 ==> 2NaNO3 + H2O
1. 1.5 what? grams? tons? bushels? I didn't get 1.5 grams.
3.
00416 what? Show your work.
Please help im sure i did it all wrong
You may not have done it wrong depending upon what the units were.
Na2O + 2HNO3 ==> 2NaNO3 + H2O
You want 0.5 mol NaNO3; how much Na2O is needed? So
0.5 mol NaNO3 x (1 mols Na2O/2 mol NaNO3) = ? molNa2O.
Then grams Na2O = mols Na2O x molar mass Na2O.
Let's go through each of the questions and determine the correct answers.
1) In order to find the mass of sodium oxide needed to react with acid to form 0.5 mol of salt, you first need to know the balanced chemical equation for the reaction between sodium oxide (Na2O) and nitric acid (HNO3). Assuming the reaction is:
Na2O + 2HNO3 -> 2NaNO3 + H2O
From the equation, you can see that one mole of sodium oxide reacts with 2 moles of nitric acid to form 2 moles of salt (NaNO3). So, if you have 0.5 mol of salt, you would need half of that amount in moles of sodium oxide, which is 0.25 mol.
To convert from moles to grams, you need to know the molar mass of sodium oxide, which is approximately 61.98 g/mol. Therefore, the mass of sodium oxide needed would be:
Mass = number of moles * molar mass
Mass = 0.25 mol * 61.98 g/mol
Mass = 15.495 g
So, the correct answer for question 1 is 15.495 grams, not 1.5.
2) To determine the amount of sodium oxide required to neutralize 230 cm^3 of a 1.0 moldm-3 nitric acid solution, you need to use the molar ratio between sodium oxide and nitric acid.
Again, referring to the balanced equation from question 1, you can see that 1 mole of sodium oxide reacts with 2 moles of nitric acid. The volume of the nitric acid solution (230 cm^3) is not directly related to the amount of sodium oxide, since they have different units.
To find the number of moles of nitric acid, you need to use the formula:
Number of moles = concentration (mol/dm^3) * volume (dm^3)
Number of moles = 1.0 moldm-3 * (230 cm^3 / 1000 cm^3/dm^3)
Number of moles = 0.23 moles
From the balanced equation, you can see that 1 mole of sodium oxide is required to react with 2 moles of nitric acid. So, to neutralize 0.23 moles of nitric acid, you would need half of that amount in moles of sodium oxide, which is 0.115 moles.
To convert from moles to grams, you can again use the molar mass of sodium oxide. The molar mass is approximately 61.98 g/mol. Therefore, the mass of sodium oxide required would be:
Mass = number of moles * molar mass
Mass = 0.115 mol * 61.98 g/mol
Mass = 7.1177 g
So, the correct answer for question 2 is approximately 7.1177 grams.
3a) To find the mass of alkali formed when 2.0 g of sodium oxide reacts with 0.05 moldm-3 of nitric acid, you can follow a similar process as in question 1.
Using the molar ratio from the balanced equation, you can find the number of moles of sodium oxide:
Number of moles = mass (g) / molar mass
Number of moles = 2.0 g / 61.98 g/mol
Number of moles = 0.0323 moles
From the balanced equation, you can see that 1 mole of sodium oxide reacts with 2 moles of nitric acid, producing 2 moles of salt. So, the number of moles of salt formed would be 0.0323 moles.
To find the mass of alkali, you need to multiply the number of moles of salt by its molar mass. The molar mass of sodium nitrate (NaNO3) is approximately 85.00 g/mol.
Mass of alkali = number of moles * molar mass
Mass of alkali = 0.0323 moles * 85.00 g/mol
Mass of alkali = 2.74655 g
So, the correct answer for part a) of question 3 is approximately 2.74655 grams.
3b) To find the concentration (moldm-3) of the alkali solution formed when all the acid has reacted, you first need to determine the volume of the alkali solution.
The volume of the nitric acid used is given as 230 cm^3, but after reacting with sodium oxide, it is not clear what volume of alkali solution will be formed. Without this information, it is not possible to determine the concentration of the alkali solution in moldm-3.
In conclusion, the correct answers are:
1) The mass of sodium oxide needed is 15.495 grams.
2) The amount of sodium oxide required to neutralize the nitric acid is approximately 7.1177 grams.
3a) The mass of alkali formed is approximately 2.74655 grams.
3b) The concentration of the alkali solution cannot be determined without the volume of the solution.