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:
2
Explanation:
There are 3 moles O2 in 1 mole CoH12O6 so 2 moles are consumed
Answer:
The first measurement is 116.84 cm, and the second measurement is 116.85 cm.
Explanation:
The problem provides both the first and second measurements so all you have to do is fill in the blank with them.
Answer:
The correct answer is option c.
Explanation:
Formula used to determine an average atomic mass :
Mass of isotope Sb-121 = 120.904 amu
Fractional abundance of Sb-121 = 57.21% = 0.5721
Mass of isotope Sb-123 = 122.904 amu
Fractional abundance of Sb-123 = 42.79% = 0.4279
Average atomic mass of Sb:
