I am having trouble with a question below:
Magnesium reacts with dilute hydrochloric acid:
I wrote the equation as:
Mg(S)+ 2H+ (aq):Mg(aq)+H2(g)
If 0.720 g of magnesium is treated with 0.950 mol L -1of hydrochloric acid calculate the:
(a) The volume of hydrocholric acid needed to react with all the magnesium
What I did for a was use the mole statment and the formula n=m
and the n= V
However I got the answer incorrect because I think I used the wrong forumals.
What are the steps to calculate the volume of HCL
No, this formula would only be for ideal gases at stp (I think, from the limited information you have given), which an aqueous solution of hydrochloric acid is not. What equations do you know that relate solution concentration to amount of substance?
You have (correctly) identified the balanced equation for the reaction. Now, how many moles of magnesium do you have? How much HCl?
Where the chlorine? Shouldn't the reaction of HCl with Mg be;
Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)
Pedantically, it's chlorid
e, but in any event it is unchanged during the reaction - have you heard of spectator ions?