The reaction ratio of hydrogen to the ammonia produced is 3:2 hence if 3 moles of hydrogen produce 2 moles of ammonia thus mathematically,
3moleH2=2mole NH3
5moleH2=?
Thus cross-multiplying
(5*2)/3= 3⅓ moles.
1) Molar mass C5H12= 5*12 +1 *12=60+12=72 g/mol
2) 40g C5H12 * 1 mol C5H12/72 g C5H12 = 40/72 mol C5H12
3) C5H12 + 8O2 ------> 5CO2 + 6H2O
by reaction 1 mol 8 mol
from problem 40/72 mol x mol
x=(40/72) * 8/1=(40*8)/72=(40)/9 mol O2
4) M(O2)=2*16 g/mol =32 g/mol
5) (40)/9 mol O2 *(32 g O2/ 1 mol )=(40 * 32)/9 =142.2 g O2
Answer:
it moves faster when heated
Answer:
4.65 L of NH₃ is required for the reaction
Explanation:
2NH₃(g) + H₂SO₄(aq) → (NH₄)₂SO₄(s)
We determine the ammonium sulfate's moles that have been formed.
8.98 g . 1mol / 132.06 g = 0.068 moles
Now, we propose this rule of three:
1 mol of ammonium sulfate can be produced by 2 moles of ammonia
Therefore, 0.068 moles of salt were produced by (0.068 . 29) / 1 = 0.136 moles of NH₃. We apply the Ideal Gases Law, to determine the volume.
Firstly we do unit's conversions:
27.6°C +273 = 300.6 K
547.9 mmHg . 1 atm / 760 mmHg = 0.721 atm
V = ( n . R . T ) / P → (0.136 mol . 0.082 L.atm/mol.K . 300.6K) / 0.721 atm
V = 4.65 L
Answer:
A) 6.00 mol.
B) 0.375 L or 375 mL
C) 6.00 M
Explanation:
Hello,
A) In this case, from the definition of molarity, we compute the moles for the given volume and concentration:
B) In this case, from the stock solution, the required volume is:
C) In this case, we apply the following formula for dilution process:
Thus, solving for the final molarity, we obtain:
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