Question #3b58a
1 Answer
Here's what I got.
Explanation:
Your starting point here will be the balanced chemical equation for this single replacement reaction
#"Mg"_text((s]) + color(red)(2)"HCl"_text((aq]) -> "MgCl"_text(2(aq]) + "H"_text(2(g]) uarr#
Notice that you have a
Now, use magnesium's molar mass to determine how many moles you have in that
#2.4 color(red)(cancel(color(black)("g"))) * "1 mole Mg"/(24.305color(red)(cancel(color(black)("g")))) = "0.09875 moles Mg"#
Next, use the molarity and volume of the hydrochloric acid solution to determine how many moles you're supplying to the reaction.
As you know, molarity tells you how many moles of solute, which in your case is hydrochloric acid, you get per liter of solution. This means that you have
#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#
This will get you
#n_(HCl) = "0.50 mol" color(red)(cancel(color(black)("L"^(-1)))) * 359 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.1795 moles HCl"#
Now, do you have enough moles of hydrochloric acid to ensure that all the moles of magnesium metal react?
Use the aforementioned mole ratio to find out. That many moles of magnesium would require
#0.09875color(red)(cancel(color(black)("moles Mg"))) * (color(red)(2)color(white)(a)"moles HCl")/(1color(red)(cancel(color(black)("mole Mg")))) = "0.1975 moles HCl"#
However, you only have
More specifically, the reaction will only consume
#0.1795color(red)(cancel(color(black)("moles HCl"))) * "1 mole Mg"/(color(red)(2)color(red)(cancel(color(black)("moles HCl")))) = "0.08975 moles Mg"#
Now, you have a
This would be the reaction's theoretical yield, i.e. what you get if all the moles of magnesium that take part in the reaction end up forming an equal number of moles of hydrogen gas, i.e. if the reaction has a
Now, you know that the reaction produced
Under these specific conditions for pressure and temperature, one mole of any ideal gas occupies exactly
Under these conditions, one mole of any ideal gas occupies
Use this to find the number of moles of hydrogen produced by the reaction - use the conversion factor
#"1 L" = 10^3"mL"#
to convert the volume of the gas from milliliters to liters.
#125.0 * 10^(-3)color(red)(cancel(color(black)("L"))) * overbrace("1 mole H"_2/(22.4color(red)(cancel(color(black)("L")))))^(color(purple)("molar volume of a gas at STP")) = "0.005580 moles H"_2#
So, the reaction should have produced
The reaction's percent yield is defined as
#color(blue)(|bar(ul(color(white)(a/a)"% yield" = "what you actually get"/"what you should theoretically get" xx 100color(white)(a/a)|)))#
Plug in your values to get
#"% yield" = (0.005580color(red)(cancel(color(black)("moles"))))/(0.08975color(red)(cancel(color(black)("moles")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)"6.2%"color(white)(a/a)|)))#
The answer is rounded to two sig figs.
If you ask me, this value is way too low, so you might want to check the values given to you for the mass of magnesium, molarity and volume of the solution, and volume of gas produced at STP.