A 1.897g sample of Mg(HCO3)2 was heated and decomposed. When the sample
1 answer:
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 %
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Answer:
0.36 M
Explanation:
There is some info missing. I think this is the complete question.
<em>Suppose a 250 mL flask is filled with 0.30 mol of N₂ and 0.70 mol of NO. The following reaction becomes possible: </em>
<em>N₂(g) +O₂(g) ⇄ 2 NO(g) </em>
<em>The equilibrium constant K for this reaction is 7.70 at the temperature of the flask. Calculate the equilibrium molarity of O₂. Round your answer to two decimal places.</em>
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Initially, there is no O₂, so the reaction can only proceed to the left to attain equilibrium. The initial concentrations of the other substances are:
[N₂] = 0.30 mol / 0.250 L = 1.2 M
[NO] = 0.70 mol / 0.250 L = 2.8 M
We can find the concentrations at equilibrium using an ICE Chart. We recognize 3 stages (Initial, Change, and Equilibrium) and complete each row with the concentration or change in the concentration.
N₂(g) +O₂(g) ⇄ 2 NO(g)
I 1.2 0 2.8
C +x +x -2x
E 1.2+x x 2.8 - 2x
The equilibrium constant (K) is:
Solving for x, the positive one is x = 0.3601 M
[O₂] = 0.3601 M ≈ 0.36 M