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:
0.1 mol
Explanation:
Number of mole= mass/molar mass
lithium have a mass number of 7 and oxygen have a mass number of 16.
so, (7x2) + 16
= 30
therefore, number of moles = 2.93/30
= 0.10
Answer:
General Formulas and Concepts:
<u>Chemistry - Gas Laws</u>
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
- Charles' Law:
Explanation:
<u>Step 1: Define</u>
Initial Volume: 5.0 L H₂ gas
Initial Temp: 273 K
Final Temp: 985 K
Final Volume: ?
<u>Step 2: Solve for new volume</u>
- Substitute:
- Cross-multiply:
- Multiply:
- Isolate <em>x</em>:
- Rewrite:
<u>Step 3: Check</u>
<em>We are given 2 sig figs as the smallest. Follow sig fig rules and round.</em>
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Answer:
d. Radon-222
Explanation:
²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He
Alpha particle is a helium nucleus with mass number 4 and atomic number 2. According to the law of conversation of mass, the sum of the mass number and atomic number must be equal on both side of the reaction.
Since the mass number of Ra is 226 and that of He is 4. The mass number of the unknown element must be 226 - 4 = 222.
Since the atomic number of Ra is 88 and that of He is 2. The atomic number of the unknown element must be 88 - 2 = 86.
Now looking in the periodic table Radon is the only element with atomic number 86.
