For the reaction,$2x$+$3y$+$4z$$ rarr$$ 5w$ Initially if 1 mole of $x$,3 mole of $y$ and 4 mole of $z$ is taken and 1.25 mole of $w$ is obtained then what is the % yield of this reaction?

Question

For the reaction,$2x$+$3y$+$4z$$ rarr$$ 5w$ Initially if 1 mole of $x$,3 mole of $y$ and 4 mole of $z$ is taken and 1.25 mole of $w$ is obtained then what is the % yield of this reaction?

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Md Ariful Islam · Answer 1 · Apr 02, 2024 09:48:30

This is a limiting reactant problem.

We know that we will need a balanced equation and moles of each reactant.

1. Gather all the information in one place with the experimental number of moles below the formulas.

$color(white)(mmmmmm)2x + 3y + 4z = 5w$
$"Amt/mol:"color(white)(ml)1color(white)(mm)3"color(white)(mm)4$
$"Divide by:"color(white)(ml)2color(white)(mm)3color(white)(mm)4$
$"Moles rxn:"color(white)(m)0.5color(white)(ml)1color(white)(mm)1$

2. Identify the limiting reactant

An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will give:

You divide the moles of each reactant by its corresponding coefficient in the balanced equation.

I did that for you in the table above.

$x$ is the limiting reactant because it gives the fewest moles of reaction.

3. Calculate the theoretical yield of $w$

$"Theoretical yield" = 1 color(red)(cancel(color(black)("mol"color(white)(l) x))) × ("5 mol " w)/(2 color(red)(cancel(color(black)("mol "x)))) = "2.5 mol"color(white)(l)w$

4. Calculate the percent yield

$"Percent yield" = "Actual yield"/"Theoretical yield" × 100 % = (1.25 color(red)(cancel(color(black)("mol"))))/(2.5 color(red)(cancel(color(black)("mol")))) × 100 % = 50 %$

For the reaction,$2x$+$3y$+$4z$$ rarr$$ 5w$ Initially if 1 mole of $x$,3 mole of $y$ and 4 mole of $z$ is taken and 1.25 mole of $w$ is obtained then what is the % yield of this reaction?

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