Answers
N₂ + 3H₂ ---> 2NH₃
stoichiometry of H₂ to N₂ is 3:1
number of H₂ moles reacted - 2.79 g / 2 g/mol = 1.40 mol
if 3 mol of H₂ reacts with 1 mol of N₂
then 1.40 mol of H₂ reacts with - 1.40/3 = 0.467 mol of N₂
mass of N₂ required - 0.467 mol x 28 g/mol = 13.1 g
mass of N₂ formed is 13.1 g
Related Questions
Write the balanced reaction and solubility product expression (KSP) for dissolving silver chromate: Ag2CrO4(s). Include all charges, stoichiometric coefficients, and phase subscripts.
Which of the characteristics below best describes organic compounds? produced by living organisms compounds of carbon and hydrogen limited to carbon compounds which are synthetic any compound containing carbon
Which response gives the products of hydrolysisofNH4Cl?A. NH4+ + HClB. NH3 + OH- + HClC. NH3 + H+D. NH4OH + HClE. No hydrolysis occurs.
What is the net ionic equation for the reaction that occurs when aqueous solutions of naoh and hno3 are mixed?
Answers
Answer:
The concentration is 0.036 mg/mL
Explanation:
Concentration = 0.2 mM = 0.2/1000 = 2×10^-4 M = 2×10^-4 mol/L × 180,000 mg/1 mol × 1 L/1000 mL = 0.036 mg/mL
Answers
Answer:
(Incomplete question)
Assuming the molarity of the weak acid is 17.4 M, the answer would be 52.4mL.
Explanation:
Equivalence point is defined as the point where moles of titrant = moles of titrand (analyte).
At equivalence point,
# moles of NaOH = # moles of weak acid
# moles of NaOH = 0.1236 × 12.43 mL
= 0.1236 mol/L × 0.01242 L
= 0.00153511 moles of NaOH
= 0.00153511 moles of weak acid.
Since the concentration of acid is not stated in your question, we will assume the concentration of the acid to be 17.4 M.
concentration = no. of moles ÷ volume
⇒ vol. = no. of moles ÷ conc.
= 0.00153511 mol ÷ 17.4 mol/L
= 0.0267 L ≈ 26.7 mL
This means that the total volume of the solution at the half equivalence point will be:
26.7 mL + 26.7 ml
= 52.4 mL.
N.B: Confirm missing variable from question: it could be concentration or volume of acid,but it is impossible to have two unknowns. Also, incase its pH of acid that's given, you can solve problem using Henderson-Hasslebauch equation.
Answers
Answer:
30.1 g NaCl
Explanation:
Your first conversion is converting grams NaOH to moles of NaOH using its molar mass (39.997 g/mol). Then, use the mole ratio of 1 mol NaCl for every 1 mol NaOH to get to moles of NaCl. Then finally multiply by the molar mass of NaCl (58.44 g/mol) to get grams of NaCl.
20.6 g NaOH • (1 mol NaOH / 39.997 g NaOH) • (1 mol NaCl / 1 mol NaOH) • (58.44 g NaCl / 1 mol NaCl) = 30.1 g NaCl
O A. flour and water
O B. sand and water
O c.
salt and water
O D.
oil and water
O E. ice and water
Answers
Final answer:
An example of a solution is salt and water.
Explanation:
An example of a solution is salt and water. When salt is mixed with water, it dissolves and forms a homogeneous mixture where the salt particles are evenly distributed throughout the water.
Learn more about solutions here:
2.05 g salicylic acid x (180g aspirin/1 mol) x (1 mol/138 g salicylic acid)
Answers
The question is incomplete; part of the data required in the question are shown:
Theoretical Yield: 2.05 g salicylic acid x (180g aspirin/1 mol) x (1 mol/138 g salicylic acid) 2. Mass of filter paper 2.56 g 3. Mass of filter paper and aspirin 5.42 g 4. Mass of aspirin (3-2) g. Percent Yield [(4)/(1)] x 100
Answer:
107%
Explanation:
We can calculate the theoretical yield as shown;
2.05g salicylic acid × 180g aspirin/1mol × 1 mol/138g of salicylic acid
Theoretical yield= 2.67 g of aspirin
Actual yield of aspirin is obtained from the experimental data;
Mass aspirin + filter paper= 5.42 g
Mass of filter paper= 2.56 g
Mass of aspirin= 5.42 g -2.56 g = 2.86 g
Hence actual yield of aspirin = 2.86 g
Percentage yield = actual yield/theoretical yield × 100
Percentage yield = 2.86/2.67 ×100 = 107%
Answers
Answer:
28.2 g of NaOH
Explanation:
We need to calculate the grams of NaOH needed to react with 25.0 g of Cl₂ in the following reaction:
2 NaOH(aq) + Cl₂(g) → NaOCI(aq + H₂0(I) + NaCl(aq)
We are going to solve this by making use of the molar ratio between Cl₂ and NaOH given by the reaction equation where we see that every mol of Cl₂ will react with 2 moles of NaOH.
So first we need to convert the 25.0 g of Cl₂ to moles:
- Molar Mass of Cl₂ = 2 x 35.45 = 70.90 g/mol
- Moles of Cl₂ = 25.0 g / 70.90 g/mol = 0.3526 moles
Then we need to calculate the moles of NaOH needed to react with these moles of Cl₂ knowing that every mol of Cl₂ will react with 2 moles of NaOH:
- moles of NaOH = 2 x moles of Cl₂ = 2 x 0.3526 moles = 0.7052 moles
Next we must convert these moles to grams:
- Molar Mass of NaOH = 22.990 + 15.999 + 1.008 = 40.00 g/mol
- Mass of NaOH = 0.7052 moles x 40.00 g/mol = 28.2 g
28.2 g are needed to react with 25.0 g of Cl₂ in the production of NaOCl