Answer:
3.50 M
Explanation:
Step 1: Write the balanced equation
4 NH₃(g) + 7 O₂(g) ⇄ 2 N₂O₄(g) + 6 H₂O(g)
Step 2: Make an ICE chart
4 NH₃(g) + 7 O₂(g) ⇄ 2 N₂O₄(g) + 6 H₂O(g)
I 0 0 3.60 3.60
C +4x +7x -2x -6x
E 4x 7x 3.60-2x 3.60-6x
Step 3: Calculate the value of x
The concentration of water at equilibrium is 0.600 M. Then,
3.60-6x = 0.600 M
x = 0.500 M
Step 4: Calculate the concentration of O₂ at equilibrium
The concentration of O₂ at equilibrium is 7x = 7(0.500M) = 3.50 M
From the question, the molarity of the desired solution was given as 0.235 M while the volume of the solution is 250 mL. Molarity is a unit of concentration used for solutions. It is necessary to first define what molarity is to be able to answer the question. Molarity is defined as number of moles of solute divided by the volume of the solution.
In this case, the amount of solid KCl required to obtain a concentration of 0.235 M in a volume of 250 mL is to be determined. The molar mass of KCl will also be used as conversion factor from unit of moles to grams. The value is 74.5513 g/mol. The following equation is used:
0.235 mol/ L x 1 L/1000 mL x 250 mL x 74.5513 g/mol KCl = 4.3799 g KCl
Thus, 4.38 g KCl is required to prepare 250 mL of 0.235 M solution.
Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
The hydrogen will react with 12 g B
Mass of B = 3.3 g H_2 × (3.6 g B/1.0 g H_2) = 12 g B
Answer: 6.98
The Relative Atomic Mass of Li = (98*7 + 2*6)/98+2
RAM of Li = (686+12)/100
RAM of Li= 6.98