Mg + 1/2 O2 → MgO
1 mol = 24 g of Mg
X mol = 12 g of Mg
x = 0.5 moles of Mg
Mg :MgO = 1:1 (coefficient from equations using mole ratio)
So
0.5 moles of MgO
1 mol MgO = (24+16) g = 40 g
0.5 moles of MgO = 0.5 × 40
= 20 g of MgO produced
The easiest way is to use the Law of Gay-Lussac. This law states that there is a direct relation between the temperature in Kelvin of a gas and the pressure.
Then, namig p the pressure and T the temperature in Kelvin and using subscripts for every state:
p/T is constant ==> p_1 / T_1 = p_2/T_2
From which you obtain:
p_2 = [p_1 / T_1] * T_2
T_1 = 33.0 + 273.15 = 306.15 K
T _2 = 21.4 + 273.15 = 294.55 K
p_1 = 1014 kPa
p_2 = 1014 kPa * 294.55 K / 306.15 K = 975.6 kPa
Answer: 1.8124 g of 3-nitrophthalhydrazide were recovered.
Explanation:
The balanced chemical reaction will be :
moles of 3-nitrophthalic acid = 
As 1 mole of 3-nitrophthalic acid gives = 1 mole of 3-nitrophthalhydrazide
0.0110 moles of 3-nitrophthalic acid gives = 
mole of 3-nitrophthalhydrazide
mass of 3-nitrophthalhydrazide = 
As the percentage yield is 93% , the mass of 3-nitrophthalhydrazide recovered = 
Therefore 1.8124 g of 3-nitrophthalhydrazide were recovered.
