When hydrochloric acid reacts with magnesium metal, hydrogen gas and aqueous magnesium chloride are produced. What volume of 1.0 M HCl is required to react completely with 4.00 g of magnesium
i did 4.00g x (mol/24.31 g Mg) x (2 mol HCl/1 mol Mg) x (L/!.0 mol HCl) = .329 L
why is this wrong?
It looks ok to me, too. Perhaps the answer should be in mL and not L. Check the problem and your typing. Make sure all of the numbers and units agree.
I don't see an error.
I just noticed that the molarity of the HCl is 1.0 M; therrefore, you have only two significant figures. Thus the answer should be reported as 0.329 = 0.33 L tp 2 s.f. That may be the problem.
thanks that worked
The calculation you performed is almost correct, but there is one error in the conversion factor. Let's break down the correct calculation step by step.
1. First, find the number of moles of magnesium (Mg) using the given mass of magnesium:
4.00 g Mg x (1 mol Mg / 24.31 g Mg) = 0.164 mol Mg
2. Next, determine the stoichiometry of the reaction between hydrochloric acid (HCl) and magnesium. From the balanced chemical equation:
Mg + 2HCl → MgCl2 + H2
It shows that 1 mol of magnesium reacts with 2 moles of hydrochloric acid.
3. Convert the number of moles of magnesium (from step 1) to the number of moles of hydrochloric acid:
0.164 mol Mg x (2 mol HCl / 1 mol Mg) = 0.328 mol HCl
Here, the moles of magnesium cancel out, leaving you with the moles of hydrochloric acid.
4. Finally, convert the number of moles of hydrochloric acid to volume using the concentration of the hydrochloric acid:
0.328 mol HCl x (1 L / 1.0 mol HCl) = 0.328 L
The conversion factor is 1 L / 1.0 mol HCl because the concentration of the hydrochloric acid is 1.0 M, which means that there is 1 mol of HCl per 1 liter of solution.
So, the correct final answer is 0.328 L of 1.0 M HCl, which is the volume required to react completely with 4.00 g of magnesium.