If we were to make room for errors, there should really be no limiting reagent because practically all of both Nitrogen and Hydrogen is used up during this reaction. If this values were actually exact, then Nitrogen would be the limiting reagent, but a very very little amount of Nitogen is needed for all the Hydrogen to react.
We solve this problem by first writing the equation
N2 + 3H2 = 2NH3
N2 = 14g*2 = 28g, 3H2 = 3(1*2) = 6g
so 28g of Nitrogen needs 6g of Hydrogen for this reaction. Thus if we had 10.67g of Hydrogen in the reaction, 6g*49.84g/28g of hydrogen is needed to react = 10.68g of Hydrogen, but since we have 10.7g of it thus it is excess and thus the limiting reagent has to be Nitrogen, but notice that 10.68g and 10.7g are practically the same, so there might actually not be a limiting reagent. Using the other value(10.7), the amount of Nitrogen required would be 10.7g*28g/6g = 49.93, and since this is slightly more than the 49.84g we have, this confirms that Nitrogen is the limiting reagent. But note still that since this values are really close, there is a possibility that there is neither a limiting nor an excess reagent
Answer:
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Answer:
Explanation:
Data:
p₁ = 694.9 mmHg; V₁ = 3.463 L
p₂ = ?; V₂ = 5.887 L
Calculation:
The <span>Anaphase stage
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Answer:
no
Explanation:
"Alkali metals are among the most reactive of all metals, which makes them suitable for specific and limited uses.
Alkali metals include lithium, sodium, potassium, rubidium, cesium and francium. These metals have large atomic radii and generally lose electrons during reactions. "
- Reference
