Answer:
Na3PO4 is excess reactant, CaCl2 is limiting reactant.
Explanation:
3CaCl2 + 2Na3PO4 ---> Ca3(PO4)2 + 6NaCl
from reaction : 3 mol 2 mol
given: 6 mol 5 mol (X)
X = (6*2)/3 = 4 mol Na3PO4
For 6 mol CaCl2 we need 4 mol Na3PO4, but we have 5 mol Na3PO4,
Na3PO4 is excess reactant, so CaCl2 is limiting reactant.
The sample has a new pressure of 274kPa. If at 105 kPa and 275K, a 220 mL sample of helium gas is contained in a cylinder with a moving piston. The sample is pushed till it has a 95.0 mL volume and 310K .
The macroscopic characteristics of ideal gases are related by the ideal gas law (PV = nRT). A gas is considered to be perfect if its particles (a) do not interact with one another and (b) occupy no space (have no volume). Where P= pressure V= volume and T = temperature.
From ideal gas equation
P₁V₁/T₁ =P₂V₂/T₂
105×220÷275 = P₂ ×95÷310
P₂= (105×220×310)÷(275×95)
P2= 7161000/26125
P2 = 274.105 kPa
Hence, the new pressure of helium gas is 274kPa
To know more about Ideas gas equation
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.90 dL is 90 mL because 1 dL is 100 mL
Answer:
Kc = 8.05x10⁻³
Explanation:
This is the equilibrium:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Initially 0.0733
React 0.0733α α/2 3/2α
Eq 0.0733 - 0.0733α α/2 0.103
We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.
Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.
3/2α = 0.103
α = 0.103 . 2/3 ⇒ 0.0686
So, concentration in equilibrium are
NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682
N₂ = 0.0686/2 = 0.0343
So this moles, are in a volume of 1L, so they are molar concentrations.
Let's make Kc expression:
Kc= [N₂] . [H₂]³ / [NH₃]²
Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³
