Question

Using Stoichiometry,calculate the mole quantity of hydrogen gas produced for each data point in Part A and Part B of this experiment using your data. Identify the concentration of HCl (part A) or mass of magnesium (part B)for when the limiting reactant changes. All calculations must be shown.

Part A

1. )0.250 g Mg

2. )0.50 M

3. )1.0 M

4. )1.5 M

5.) 2.0 M

6.) 2.5 M

7.) 3.0 M

8.) 3.5 M

PART B

1. 2.0 M HCl -Entre moles of calculated moles of H2 (g) from Part A.

2. 0.050 g

3. 0.10 g

4. 0.15 g

5. 0.20 g

6. 0.25 g -enter moles of calculated moles of H2 (g) from Part A.

7. 0.30 g

8. 0.35 g

9. 0.40 g

http://tinypic.com/r/kah0rt/9 Here is the data page. I hope the data page is legible!

PLEASE HELP! Even if you could just explain the process for just one of them that would be helpful. I assume once you get the first one the other ones should follow the same method of solving. I would really appreciate anyone's help as soon as possible.

Answer 1

The reaction is

Mg(s) + 2HCl --> MgCl2 + H2(g)

so each mole of Mg will react with two moles of HCl to give one mole of H2.

In each reading we have to calculate

Moles of Mg = Mass of metal / atomic weight

Moles of HCl = Molarity of HCl x volume of HCl / 1000

Limiting reagent

Moles of H2 produced (based on limiting reagent)

Volume of H2 = 22.4 X moles of H2 produced(assuming the system to be at STP, where volume occupied by each mole of gas is 22.4 L)

Mass of metal	atomic weight	moles of Mg	Volume of acid	Molarity of HCl	Moles of HCl	Limiting reagent	Moles of Gas	Volume of Gas (L)volume of Gas (mL)	volume of Gas (mL
0.2510	24	0.0105	10.0000	0.5000	0.0050	HCl	0.0025	0.0560	56.00
0.2510	24	0.0105	10.0000	1.0000	0.0100	HCl	0.0050	0.1120	112.00
0.2500	24	0.0104	10.0000	1.5000	0.0150	HCl	0.0075	0.1680	168.00
0.2560	24	0.0107	10.0000	2.0000	0.0200	HCl	0.0100	0.2240	224.00
0.2500	24	0.0104	10.0000	2.5000	0.0250	Mg	0.0104	0.2333	233.33
0.2510	24	0.0105	10.0000	3.5000	0.0350	Mg	0.0105	0.2343	234.27
0.2540	24	0.0106	10.0000	3.0000	0.0300	Mg	0.0106	0.2371	237.07