The theoretical molar yield of lead (II) chloride will be 9 moles.
<h3>Stoichiometric calculation</h3>
First, we need to look at the equation of the reaction:
From the equation, the 1 mole of Pb2+ ion requires 2 moles of Cl- ion in order to produce 1 mole of lead (II) chloride.
Thus, with 18 moles Cl- ions, 9 moles of Pb2+ would be needed, instead of 12 moles. Pb2+ is simply in excess while Cl- can be said to be limiting.
Therefore, Cl- will determine how much of lead (II) chloride that will be produced. The ratio is 2 moles of Cl- to 1 mole of lead (II) chloride.
With 18 moles Cl-, 9 moles of lead (II) chloride will, thus, be produced.
More on mole ratios can be found here: brainly.com/question/14425689
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Answer:
option (b)
Explanation:
A gases can be compressed much more than a solid or liquid because the particles of a gas are very far apart
Answer:
Explanation:
4NH₃ + 3O₂- --> 2N₂ + 6 H₂O
4 mole 3 moles 2 moles 6 moles
4 moles of NH₃ reacts with 3 moles of oxygen
1.37 moles of NH₃ reacts with 3 x 1.37 / 4 mols of oxygen
corresponding amount of O₂ = 3 x 1.37 / 4 mols
= 1.0275 moles
corresponding amount of N₂ = 2 x 1.37 / 4 moles
= .685 moles
corresponding amount of H₂O = 6 x 1.37 / 4 moles
= 2.055 moles
Given amount of product N₂ = 3.82 moles
corresponding amount of NH₃ = 4 x 3.82 / 2 moles
= 7.64 moles
corresponding amount of O₂ = 3 x 3.82 / 2 moles
= 5.73 moles
corresponding amount of H₂O = 6 X 3.82 / 2 moles
= 11.46 moles .
<span>Answer: 0.094%
</span><span>Explanation:
</span>
<span></span><span /><span>
1) Equilibrium chemical equation:
</span><span />
<span>Only the ionization of the formic acid is the important part.
</span><span />
<span>HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq).
</span><span />
<span>2) Mass balance:
</span><span />
<span> HCOOH(aq) HCOO⁻(aq) H⁺(aq).
Start 0.311 0.189
Reaction - x +x +x
Final 0.311 - x 0.189 + x x
3) Acid constant equation:
</span><span />
<span>Ka = [HCOO-] [N+] / [HCOOH] = (0.189 + x) x / (0.311 -x)
</span><span />
<span>= (0.189 + x )x / (0.311 - x) = 0.000177
4) Solve the equation:
You can solve it exactly (it will lead to a quadratic equation so you can use the quadratiic formula). I suggest to use the fact that x is much much smaller than 0.189 and 0.311.
</span><span />
<span>With that approximation the equation to solve becomes:
</span><span>0.1890x / 0.311 = 0.000177, which leads to:</span>
<span /><span>
x = 0.000177 x 0.311 / 0.189 = 2.91 x 10⁻⁴ M
5) With that number, the percent of ionization (alfa) is:
</span><span />
<span>percent of ionization = (moles ionized / initial moles) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (concentration of ions / initial concentration) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (0.000291 / 0.311)x 100 = 0.0936% = 0.094%
</span>
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