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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 =
x 100
putting the values in the equation:
percent yield = 
= 0.015 x100
PERCENT YIELD = 1.5 %
If we consider a combustion reaction of Methane:
The balanced equation is:
CH4 + 2O2 ---> 2H2O + CO2
The rate of appearance of H2O is rH2O, rate of disappearance of O2 is -rO2
(rH2O)^2 = (-rO2)^2
rH2O = -rO2
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Answer:
Define a problem, form a hypothesis, gather experimental data, form a conclusion