What is the mass in grams of 0.280 mole sample of sodium hydroxide NaOH
1 answer:
Answer:
The mass of 0.280 mole sample of sodium hydroxide NaOH is 11.2 grams.
Explanation:
To know the mass in grams of 0.280 moles of sample of sodium hydroxide NaOH, you must know the molar mass of the compound, that is, the mass of one mole of a substance, which can be an element or a compound.
So you know:
- Na: 23 g/mole
- O: 16 g/mole
- H: 1 g/mole
So, the molar mass of NaOH is:
NaOH= 23 g/mole + 16 g/mole+ 1 g/mole= 40 g/mole
Then the following rule of three can be applied: if in 1 mole of sodium hydroxide there are 40 grams, in 0.280 moles how much mass is there?
mass= 11.2 grams
<u><em>The mass of 0.280 mole sample of sodium hydroxide NaOH is 11.2 grams.</em></u>
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Answer:
a) HCl is the limiting reagent.
b) Mass of NH₄Cl formed = 6.68 g
c) Pressure of the gas remaining in the flask = 1.742 atm
Explanation:
The complete Question is presented in the attached image to this solution.
To solve this question, we first need to obtain the limiting regaent for this reaction.
The limiting reagent is the reagent that is in short supply in the reaction and is used up in the reaction. It determines the amount of products that will be formed and the amount of other reactants that will be required for the reaction.
NH₃ (g) + HCl (g) ⟶ NH₄Cl (s)
1 mole of NH₃ reacts with 1 mole of HCl
we first convert the masses of the gases available to number of moles.
Number of moles = (Mass/Molar Mass)
Molar mass of NH₃ = 17.031 g/mol, Molar mass of HCl = 36.46 g/mol
Number of moles of NH₃ = (4.55/17.031) = 0.2672 mole
Number of moles of HCl = (4.55/36.46) = 0.1248 mole
Since 1 mole of NH₃ reacts with 1 mole of HCl
It is evident that HCl is in short supply and is the limiting reagent.
NH₃ is in excess.
So, to calculate the amount of NH₄Cl formed,
1 mole of HCl gives 1 mole of NH₄Cl
0.1248 mole of HCl will also gove 0.1248 mole of NH₄Cl
Mass (Number of moles) × (Molar Mass)
Molar mass of NH₄Cl = 53.491 g/mol
Mass of NH₄Cl formed = 0.1248 × 53.491 = 6.68 g
c) The gas remaining in the flask is NH₃
0.1248 mole of NH₃ is used up for the reaction, but 0.2672 mole was initially available for reaction,
The amount of NH₃ left in the reacting flask is then
0.2672 - 0.1248 = 0.1424 mole.
Using the ideal gas Equation, PV = nRT
We can obtain the rrequired pressure of the remaining gas in the flask
P = Pressure = ?
V = Volume = 2.00 L
n = number of moles = 0.1424 mole
R = molar gas constant = 0.08205 L.atm/mol.K
T = absolute temperature in Kelvin = 25 + 273.15 = 298.15 K
P = (nRT/V)
P = (0.1424×0.08205×298.15/2) = 1.742 atm
Hope this Helps!!!
