In this reaction: Mg (s) + I₂ (s) → MgI₂ (s) If 2.68 moles of Mg react with 3.56 moles of I₂, and 1.76 moles of MgI₂ form, what
1 answer:
Answer:
Explanation:
Hello,
In this case, for the given chemical reaction, we first identify the limiting reactant by noticing that due to the 1:1 mole ratio for magnesium to iodine the reacting moles must the same, nevertheless, there are only 2.68 moles of magnesium versus 3.56 moles of iodine, for that reason, magnesium is the limiting reactant, so the theoretical turns out:
Thus, we compute the percent yield as:
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Answer:
a. HCl.
b. 0.057 g.
c. 1.69 g.
d. 77 %.
Explanation:
Hello!
In this case, since the reaction between magnesium and hydrochloric acid is:
Whereas there is 1:2 mole ratio between them.
a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:
Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.
b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:
c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:
Thus, the mass of excess magnesium turns out:
d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:
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