Answer:
a
Explanation:
it would be the most reasonable
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.
Answer:
0.0165 (L) * 0.750 (mol/L) = 0.0124 mol NaOH in 16.5 mL
Explanation:
The question here is solved using basic chemistry. CaCl2(aq) is an ionic compound which will have the releasing of 2 Cl⁻ ions ions in water for every molecule of CaCl2 that dissolves.
CaCl2(s) --> Ca+(aq) + 2 Cl⁻(aq)
[Cl⁻] = 0.65 mol CaCl2/1L × 2 mol Cl⁻ / 1 mol CaCl2 = 1.3 M
The answer to this question is [Cl⁻] = 1.3 M
