If 35.8 g Mg react with 82.3 g HCl according to the reaction below, how many grams of hydrogen gas will be produced, and how many grams of the excess reactant will be left over? Unbalanced equation: Mg + HCl → MgCl2 + H2 Be sure to show all of your work.
@dude @hero
start with the equation Mg + HCl → MgCl2 + H2 notice how it is unbalanced (you have 2 Cl on the right side and 1 Cl on the left side; you also have 1 H on the left side and 2H on the right side) thankfully, we can simply add a 2 in front of the HCl to get both sides balanced Mg + 2HCl → MgCl2 + H2
now, we need to find out which reactant (Mg or HCl) is limiting (a.k.a which reactant limits how much product we have) to do this, we figure out whether Mg or HCl produces less product (I will use H2 for my demonstration.) setting up the stoich: \[\frac{ 35.8g~Mg }{ }\frac{ 1~mol~Mg }{ 24.31g }\frac{ 1~mol~H2 }{ 1~mol~Mg }\] \[\frac{ 82.3 g~HCl }{ }\frac{ 1~mol~HCl}{ 36.46g }\frac{ 1~mol~H2 }{ 2~mol~HCl }\] perform the calculations to see which one is limiting (should be the smaller result)
Dang vocaloid good job
take your limiting reactant, and convert this to grams H2 to find out how much H2 is produced. (since we calculated moles H2 previously, you can simply multiply the previous calculation by the molar mass of H2 to get to grams H2). that will answer the first part of the question
to address the second part of the question, we want to know how much of the non-limiting reactant is left. to do this, take the limiting reactant and convert this to the other reactant with the appropriate stoich. that will tell you how much of the non-limiting reactant is used.. then simply subtract (original amount of the non-limiting reactant) - (amount of non-limiting reactant) to get the amt. of excess reactant leftover, thus solving the second part of the question I know this is a lot, but please review the steps and let me know if anything I said was confusing.
thank you so much!
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