Consider the reaction of magnesium metal with hydrochloric acid to produce magnesium chloride and hydrogen gas. If 3.75 mol of magnesium and 3.75 mol of hydrochloric acid are reacted, how many grams of excess reactant are left over?
a. 1.88 g
b. 91.2 g
c. 68.4 g
d. g
e. 45.6 g
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To find the excess reactant, we need to determine which reactant is limiting.
First, we can calculate the amount of magnesium chloride produced by converting the moles of magnesium to moles of magnesium chloride:
3.75 mol Mg x (1 mol MgCl2 / 1 mol Mg) = 3.75 mol MgCl2
Next, we can calculate the amount of hydrochloric acid needed to produce 3.75 moles of magnesium chloride:
3.75 mol MgCl2 x (2 mol HCl / 1 mol MgCl2) = 7.5 mol HCl
Since we have 3.75 mol of hydrochloric acid, it is the limiting reactant.
Now, we can calculate the amount of magnesium chloride produced by 3.75 moles of hydrochloric acid:
3.75 mol HCl x (1 mol MgCl2 / 2 mol HCl) = 1.875 mol MgCl2
The amount of magnesium chloride produced is 1.875 mol.
To find the excess reactant, we subtract the amount of magnesium chloride produced from the amount of hydrochloric acid used:
3.75 mol HCl - 1.875 mol MgCl2 = 1.875 mol HCl
To convert this to grams, we use the molar mass of hydrochloric acid:
1.875 mol HCl x 36.46 g/mol = 68.4 g
Therefore, the answer is c. 68.4 g
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