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
