Answer:
Explanation:
a) The mass of the reactants is 2.36 grams, and the mass of the products is 1.57 grams plus the mass of the carbonic acid. Thus, using the law of conservation of mass, we get the mass of the carbonic acid is 2.36 - 1.57 = 0.79 grams.
b) The gram-formula mass of sodium bicarbonate is 84.006 g/mol, meaning that 2.36/84.006 = 0.028 moles were consumed. Thus, this means that in theory, 0.014 moles of carbonic acid should have been produced, which would have a mass of (0.014)(62.024)=0.868 grams. Thus, the percentage yield is (0.79)/(0.868) * 100 = 91%
Answer:
3 : 1
Explanation:
Let the rate of He be R1
Molar Mass of He (M1) = 4g/mol
Let the rate of O2 be R2
Molar Mass of O2 (M2) = 32g/mol
Recall:
R1/R2 = √(M2/M1)
R1/R2 = √(32/4)
R1/R2 = √8
R1/R2 = 3
The ratio of rate of effusion of Helium to oxygen is 3 : 1
Reaction equation:
Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O
Moles of Al(OH)₃:
moles = mass/Mr
= 1.51 / (27 + 17 x 3)
= 0.019
Molar ratio Al(OH)₃ : HCl = 1 : 3
Moles of HCl required = 0.019 x 3
=0.057
concentration = moles/volume
volume = 0.057 / 0.1
= 0.57 dm³
= 570 ml
N₂O
N = 14.01 amu
O ≈ 16 amu
14.01 × 2 = 28.02 ; 2 Nitrogen
44.02 - 28.02 = 16 ; 1 Oxygen
Check
28.02 + 16 ≈ 44.02amu
