The given thermochemical reaction is between hydrogen gas and chlorine gas to form hydrogen chloride.
This can be represented as:
Δ
=-184.6 kJ/mol
So when two moles of HCl is formed, 184.6 kJ of energy is released.
Calculating the heat released when 3.18 mol HCl (g) is formed in the reaction:
Therefore, 293.5 kJ of heat is released when 3.18 mol HCl is formed in the reaction between hydrogen and chlorine.
A common factor is low pressure system.
In this question, you are given the NaOH volume but asked for concentration.
Don't forget that for every 1 mol of NaOH there will be 1 mol OH- ion, but for every 1 mol of H2SO4 there will be 2 mol of H- ion.
To neutralize you need the same amount of OH- and H+, so the equation should be:
OH-= H+
<span>35.50cm3 * x*1= 25cm3* 0.2mol/dm3 *2
</span>x= 10/35.5 mol/dm3= 0.2816/dm3
Answer:
Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.
Explanation:
- It is a stichiometry problem.
- We should write the balance equation of the mentioned chemical reaction:
<em>2Al + 3CuCl₂ → 3Cu + 2AlCl₃.</em>
- It is clear that 2.0 moles of Al foil reacts with 3.0 moles of CuCl₂ to produce 3.0 moles of Cu metal and 2.0 moles of AlCl₃.
- Also, we need to calculate the number of moles of the reported masses of Al foil (0.50 g) and CuCl₂ (0.75 g) using the relation:
<em>n = mass / molar mass</em>
- The no. of moles of Al foil = mass / atomic mass = (0.50 g) / (26.98 g/mol) = 0.0185 mol.
- The no. of moles of CuCl₂ = mass / molar mass = (0.75 g) / (134.45 g/mol) = 5.578 x 10⁻³ mol.
- <em>From the stichiometry Al foil reacts with CuCl₂ with a ratio of 2:3.</em>
∴ 3.85 x 10⁻³ mol of Al foil reacts completely with 5.578 x 10⁻³ mol of CuCl₂ with <em>(2:3)</em> ratio and CuCl₂ is the limiting reactant while Al foil is in excess.
- From the stichiometry 3.0 moles of CuCl₂ will produce the same no. of moles of copper metal (3.0 moles).
- So, this reaction will produce 5.578 x 10⁻³ mol of copper metal.
- Finally, we can calculate the mass of copper produced using:
mass of Cu = no. of moles x Atomic mass of Cu = (5.578 x 10⁻³ mol)(63.546 g/mol) = 0.354459 g ≅ 0.36 g.
- <u><em>So, the answer is:</em></u>
<em>Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.</em>
The chemical reaction would be written as
2 AsF3<span> + 3 CCl4 = 2 AsCl3 + 3 CCl2F2
</span>
We use the given amounts of the reactants to first find the limiting reactant. Then use the amount of the limiting reactant to proceed to further calculations.
150 g AsF3 ( 1 mol / 131.92 g) = 1.14 mol AsF3
180 g CCl4 (1 mol / 153.82 g) = 1.17 mol CCl4
Therefore, the limiting reactant would be CCl4 since it would be consumed completely. The theoretical yield would be:
1.17 mol CCl4 ( 3 mol CCl2F2 / 3 mol CCl4 ) = 1.17 mol CCl2F2
