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
Dang that’s crazy.. Goodluck ..
Answer:
This is all true if the atom has to be neutral.
Also what does V mean?
Helium: one shell with 2 neutrons and 2 protons in the center, with 2 electrons in the first shell.
Lithium: two shells with 4 neutrons and 3 protons in the center, with 2 electrons in the first shell, and 1 electron in the second shell.
Nitrogen: two shells with 7 neutrons and 7 protons in the center, with 2 electrons in the first shell, and 5 electrons in the second shell.
Flourine: two shells with 9 protons and 10 neutrons in the center, with 2 electrons in the first shell, and 7 electrons in the second shell.
Neon: two shells with 10 neutrons and 10 protons in the center, with 2 electrons in the first shell, and 8 electrons in the second shell.
Boron: two shells with 6 neutrons and 5 protons in the center, with 2 electrons in the first shell, and 3 electrons in the second shell.
Answer:
275g
Explanation:
Depending on the molar mass you are given, you can use that to solve this.
(I'm going based on my science class' molar mass of sulphur being 32.07g/mol)
Starting off, the formula for finding moles is
n=m/M (moles = mass / molar mass)
We can manipulate this equation to solve for mass.
m=Mn
now fill in what we now.
m = 32.07*8.56
mass = 274.5192
Now round for significant digits (if you are needed to do)
mass = 275g