Answer:
The molar mass of the unknown gas is 154.4 g/mol
Explanation:
Step 1: Data given
It takes 986 seconds for an unknown gas to effuse
It takes 811 seconds for chlorotrifluoromethane (CClF3) gas to effuse
Molar mass of chlorotrifluoromethane (CClF3) = 104.46 g/mol
Step 2: Calculate the molar mass of the gas
t1/t2/ √(M1/M2)
⇒with t1 = the time needed for the unknown gas to effuse = 986 seconds
⇒with t2 = the time needed for CClF3 to effuse = 811 seconds
⇒with M1 = the molar mass of the unknown gas = TO BE DETERMINED
⇒with M2 = the molar mass of CClF3 = 104.46 g/mol
986/811 = √(M1/104.46)
(986/811)² = M1 / 104.46
M1 = 154.4 g/mol
The molar mass of the unknown gas is 154.4 g/mol
Answer:
162 kJ
Explanation:
The reaction given by the problem is:
- NBr₃(g) + 3 H₂O(g) → 3 HOBr(g) + NH₃(g) ∆H = +81 kJ
If we turn it around, we have:
- 3 HOBr(g) + NH₃(g) → NBr₃(g) + 3 H₂O(g) ∆H = -81 kJ
If we think now of HOBr and NH₃ as our reactants, then now we need to find out <u>which one will be the </u><em><u>limiting reactant</u></em> when we have 9 moles of HOBr and 2 moles of NH₃:
- When we have 1 mol NH₃, we need 3 mol HOBr. So when we have 2 moles NH₃, we need 6 moles HOBr. We have more than 6 moles HOBr so that's our <em>reactant in excess</em>, thus NH₃ is our limiting reactant.
-81 kJ is our energy change when there's one mol of NH₃ reacting, so we <u>multiply that value by two when there's two moles of NH₃ reacting</u>. The answer is 81*2 = 162 kJ.
Answer: 44,8 l. of CO2 and 72 g. of water will be produced
Explanation:
Answer:
We need 1.714 moles N2
Explanation:
Step 1: Data given
The reaction yield = 87.5 %
Number of moles NH3 produced = 3.00 moles
Step 2: The balanced equation
N2(g)+ 3H2(g) →2NH3(g)
Step 3: Calculate moles N2
For 2 moles NH3 produced we need 1 mol N2 and 3 moles H2
This means, if the yield was 100%, for 3.00 moles NH3 produced , we need 1.5 moles N2
For a 87.5 % yield:
we need more N2, increased by a ratio of 100/87.5.
100/87.5 * 1.5 = 1.714 moles N2
Answer: You would separate it by dissolving them.
