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sdas [7]
3 years ago
13

A 1.897g sample of Mg(HCO3)2 was heated and decomposed. When the sample

Chemistry
1 answer:
svlad2 [7]3 years ago
8 0

1.5 % is the percent yield in the reaction.

Explanation:

Given that:

original mass of the sample used in reaction = 1.897 grams

product formed after decomposition = 1.071 grams

The reaction for the decomposition:

Mg(HCO3)2 (s) ⇒ CO2 (g) + H2O (g) + MgCO2 (s)

It says that 1 mole of Mg(HCO3)2  yielded 1 mole of  MgCO2  on decomposition

68.31 grams/mole or 68.31 grams of MgCO2 is formed

percent yield = \frac{actual yield}{theoretical yield} x 100

putting the values in the equation:

percent yield = \frac{1.071}{68.31}

                       = 0.015 x100

   PERCENT YIELD = 1.5 %

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How many grams of Lead(II) nitrate are present in 150.0 mL of an Lead(II) nitrate solution that is 0.07268 M?
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