Question

What is the % yield when 1.5 grams of LT2 are produced from 2.6 moles of MT and 1.4 g of L(OH)₂? [Molar mass of L(OH)2-54 g/mol and LT2-80 g/mollBalanced Reaction: 2 MT+L(OH)₂ -->2 MOH + LT₂A. % YieldB: The unit and compound of the THEORETICAL YIELD is

Answer 1

Step 1

The reaction must be written, balanced, and completed:

2 MT + L(OH)2 => 2 MOH + LT2

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Step 2

Information provided:

Actual yield = 1.5 g LT2

2.6 moles MT

1.4 g L(OH)2

Molar masses of:

L(OH)2 = 54 g/mol

LT2 = 80 g/mol

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Step 3

The limithing reactant:

By stoichiometry,

2 MT + L(OH)2 => 2 MOH + LT2

2 moles MT ----------- 54 g L(OH)2

2.6 moles MT ----------- X

X = 2.6 moles MT x 54 g L(OH)2/2 moles MT

X = 70.2 g L(OH)2

For 2.6 moles of MT, 70.2 g of L(OH)2 is needed but there is only 1.4 g of L(OH)2, so the limiting reactant is L(OH)2

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Step 4

The theroretical yield: (units of it = g, like the actual yield)

By stoichiometry,

2 MT + L(OH)2 => 2 MOH + LT2

54 g L(OH)2 -------------- 80 g LT2

1.4 g L(OH)2 -------------- X

X = 2.07 g LT2 = the theoretical yield

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Step 5

% yield is defined as follows:

`\begin{gathered} \%yield\text{ = }\frac{Actual\text{ yield}}{Theoretical\text{ yield}}x100 \\ \%yield\text{ = }\frac{1.5\text{ g LT2}}{2.07\text{ g LT2}}x100\text{ = 72.5\%} \end{gathered}`

Answer: %yield = 72.5 %