Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Answer:
Alpha decay is one type of radioactive decay, in which an atomic nucleus emits an alpha particle, and thereby transforms (or "decays") into an atom with a mass number decreased by 4 and atomic number decreased by 2.
Explanation:
The answer is B. The earth's axis is tilted so that the Northern Hemisphere faces the sun. Hope this helps and good luck! :)
Answer:
the mass number of the atom or ion is 6
<u>Answer</u>:
mass of magnesium oxide is 4.2 grams.
<u>Explanation</u>:
Balanced equation: 2Mg + O2 → 2MgO.
Given that 2.5 g Mg reacts with 3.0 g O2
- mass = moles * Mr
- moles = mass/Mr
- Mr = mass/moles
First find the moles of Mg:
moles = mass/Mr
moles = 2.5/ 24
moles = 0.104 moles
Using molar ratio, magnesium oxide will have same moles as Mg as they both have "2" molar ratio. So moles of Mg is 0.104 moles.
mass = moles * Mr
mass = 0.104 * 40
mass - 4.1667 g
mass = 4.2 g ___________this is the mass of MgO