Answer:
Limiting reactant is NiSO₄
Explanation:
The reaction of aluminum metal with aqueous nickel(II) sulfate to produce aqueous aluminum sulfate and nickel is:
2 Al(s) + 3 NiSO₄ → Al₂(SO₄)₃ + 3 Ni
<em>That means 2 moles of Al react with 3 moles of nickel sulfate.</em>
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Moles of Al and NiSO₄ are:
Al: 108g × (1mol / 26.98g) = 4.00 moles of Al
NiSO₄: 464g × (1mol / 154.75g) = 3.00 moles of NiSO₄
For a complete reaction of aluminium there are necessary:
4.00mol Al ₓ ( 3 moles NiSO₄ / 2 moles Al) = 6 moles of NiSO₄
As you have just 3.00 moles of NiSO₄, the <em>limiting reactant is NiSO₄</em>
This is molarity: moles of solute/liters of solution. (Not molality)
1. Plug in what we know:
500 mL = 0.5 L
0.80 = moles/0.5
0.80*0.5 = moles
moles = 0.4
2. NaOH is given as 40 g/mole, so calculate the grams:
0.4 * 40 = 16 grams
answer: 16 grams
Answer: 8.28g Na
Explanation: use ideal gas law
PV= nRT
Solve for moles of Cl2
n= PV/ RT
Substitute:
= 1 atm x 4.0 L / 0.08205 L.atm/ mol. K x 273 K
= 0.18 moles Cl2
Do stoichiometry to solve for m of Na
2 Na + Cl2 => 2 NaCl2
=0.18 moles Cl2 x 2 mol Na/ 1 mol Cl2 x 23g Na / 1 mol Na
= 8.28 g Na.
Answer:
diffraction
Explanation:
the correct answer is diffraction
You will want to find how many grams are in a whole mole so you know which element it is. To do this, find out how much of a mole you have.
4.95 x 10^23 atoms / 6.022 x 10^23 atoms (one whole mole of any element) = .8219860511 or ~82% of 1 mole
Now we know that, find what to multiply 20 g by to get the rest of the mole.
1 mole / .8219860511 mole = 1.216565657
20 g x 1.216565657 = ~24.33 g / mol
Now that you have grams per mole, you can look at the periodic table and the molar masses to see which this number is closely aligned.
Your answer is Magnesium (Mg), which has a molar mass of 24.305 g
