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faltersainse [42]
3 years ago
5

What is the percent yield of this reaction if 22.o g of Mgl2 is produced by the reaction of 25.0 g of Mg with 25.0 g of l2?

Chemistry
1 answer:
expeople1 [14]3 years ago
8 0

% yield = 80.719

<h3>Further explanation</h3>

Given

22.0 g of Mgl₂

25.0 g of Mg

25.0 g of l₂

Required

The percent yield

Solution

Reaction

Mg + I₂⇒ MgI₂

mol Mg = 25 g : 24.305 g/mol = 1.029

mol I₂ = 25 g : 253.809 g/mol = 0.098

Limiting reactant = I₂

Excess reactant = Mg

mol MgI₂ based on I₂, so mol MgI₂ = 0.098

Mass MgI₂ (theoretical):

= mol x MW

= 0.098 x 278.114

= 27.255 g

% yield = (actual/theoretical) x 100%

% yield = (22 / 27.255) x 100%

% yield = 80.719

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The pressure of the gas in the flask (in atm) when Δh = 5.89 cm is 1.04 atm

<h3>Data obtained from the question</h3>

The following data were obtained from the question:

  • Atmospheric pressure (Pa) = 730.1 torr = 730.1 mmHg
  • Change in height (Δh) = 5.89 cm
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<h3>How to determine the pressure of the gas</h3>

The pressure of the gas can be obtained as illustrated below:

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