Percentage yield = (actual yield / theoretical
yield) x 100%<span>
The balanced equation for the reaction is,
CH₄(g) + Cl₂<span>(g) </span>→ CH₃Cl(g)
+ HCl(g)</span><span>
Since there is excess of Cl₂ gas, we can assume that all of CH₄ gas are reacted.</span><span>
Moles of CH₄(g) = mass / molar mass</span><span>
= 25.0 g / 16 g/mol
= 1.5625 mol
The stoichiometric ratio between CH₄(g) and CH₃Cl(g)
is 1 : 1</span><span>
Hence moles of CH₃Cl(g) = 1.5625 mol</span><span>
Molar mass of CH₃Cl(g) = 50.5 g/mol</span><span>
Mass of CH₃Cl(g) = number of moles x molar mass</span><span>
= 1.5625 mol x 50.5 g/mol
<span>
= 78.9 g</span>
Hence theoretical yield = 78.9 g
Actual yield = 45.0 g
Hence,
<span> Percentage yield = (45.0 g / 78.9 g) x 100% </span>
<span>
= 57.03%</span></span>
Explanation:
3Ca(s) + 2AlCl3(aq) -> 3CaCl2(aq) + 2Al(s)
According to the question, Ca is the limiting reactant.
Therefore, we equate Ca to Aluminium which is the product whose mass we want to find
Molar mass of Ca- 40g/mol
". ". of Al- 27g/mol
3Ca --> 2Al
3×40 --> 2×27
9.2 --> x
x = 9.2×2×27= 496.8÷120=4.14
Mixture between a liquid and particles of a solid
Density is mass over volume
Density = 106g/23cm³ = 4.6g/cm³
