B.Mg(OH)2 (the active ingredient in milk of magnesia)
Answer:
F2 is the limiting reactant
27.6 grams of NaF is produced.
Explanation:
Balance the equation first.
2Na+ F2 ---> 2NaF
To find the limiting reactant, solve for how much NaF can be produced with Na and F2
12.5g F2 x (1 mole F2/ 38.00 grams F2)x (2 mole NaF/ 1 mole F2)
=0.658 moles NaF
16.2g Na x (1 mole Na/ 22.99 grams Na)x (2 mole NaF/ 2 mole Na)
=0.705 moles NaF
Since F2 produced the least NaF, F2 is the limiting reactant.
Now, to find how much NaF there is, use the moles solved above with F2 as the limiting reactant.
0.658 moles NaF x (41.99 grams NaF/ 1 mole NaF)= 27.6 moles NaF
27.6 moles of NaF would be theoretically produced.
Is there answer choices ??
Noble gases have complete valence electron shells
Answer:
5.66 %.
Explanation:
<em>mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.</em>
<em />
<em>mass % = (mass of solute/mass of solution) x 100.</em>
<em></em>
mass of potassium nitrite = 30.0 g,
mass of the solution = mass of water + mass of potassium nitrite = 500.0 g + 30.0 g = 530.0 g.
<em>∴ mass % = (mass of solute/mass of solution) x 100</em> = (30.0 g/530.0 g) x 100 = <em>5.66 %.</em>
