Question

You have a 3.0 g of solid magnesium metal in 250 ml of 1.0mol/l hydrochloric acid solution that is in the beaker. you place a large gas collecting tube over the magnesium and vollect the product, hydrogen gas. assume all the conditions at the lab are SATP.
a) write a chemical balanced equation
b) calculate the number of theoretical moles of hydrogen that should be reduced in this reaction.
c) calculate the volume of hydrogen gas produced. remember that the gas is collected over water.
d) you now quickly do a burning splint test on the hydrogen. jt reacts with just enough oxygen in the gas collecting tube to produce liquid water. what mass of liquid water is produced? (new balanced equation)​

Answer 1

Answer:

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

b) 0.125 moles of hydrogen gas is reduced in the reaction.

C) 3.18 L

d)2.25 g of water

Explanation:

a) the equation of this reaction is;

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

b)

Number of moles= concentration × volume= 1.0 × 250/1000 = 0.25moles of HCl

From the equation;

2 moles of HCl yields 1 mole of hydrogen gas

Hence 0.25 moles of HCl yields 0.25 × 1/2 = 0.125 moles of hydrogen gas

Thus 0.125 moles of hydrogen gas is reduced in the reaction.

c)

P= 760 mmHg (standard pressure)

V= ????

T= 298 K

n= 0.125 moles

R= 0.082 atm dm-3K-1mol-1

Since the gas is collected over water, SVP of hydrogen at 25°c is 28mmHg

Therefore; P=760-28= 732mmHg

But

1 atm =760 mmHg

Therefore 732 mmHg= 732/760= 0.96 atm

PV=nRT

V= nRT/P

V= 0.125 × 0.082 × 298/0.96

V= 3.18 L

Note 1dm-3=1L

d)

2H2(g) + O2(g) ----> 2H2O(g)

From the equation;

2 moles of hydrogen yields 2 moles of water

0.125 moles of hydrogen yields 0.125 moles of water

Mass of water = 0.125 moles × 18gmol-1 = 2.25 g of water