Answer:
Amount of excess Carbon (ii) oxide left over = 23.75 g
Explanation:
Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂
Molar mass of Fe₂O₃ = 160 g/mol;
Molar mass of Carbon (ii) oxide = 28 g/mol
From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g) of carbon (ii) oxide
450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide
Therefore the excess reactant is carbon (ii) oxide.
Amount of excess Carbon (ii) oxide left over = 260 - 236.25
Amount of excess Carbon (ii) oxide left over = 23.75 g
Answer:
the molarity is 3.68 moles/L
Explanation:
the molality of the solution of sucrose is
m= moles of glucose / Kg of solvent (water)= 6.81 ,
since the molecular weight of glucose is 180.156 gr/mole , then per each kilogram of solvent there is
6.81 moles*180.156 gr/mole + 1000 gr of water = 2226.86 gr of solution
from the density
volume of solution = mass of solution/density = 2286.86 gr / 1.2 gr/ml = 1855.71 ml
therefore there is 1000 gr of water in 1855.71 ml
then the molarity M is
M= moles of glucose / L of solution = (moles of glucose / Kg of solvent) * (Kg of solvent/L of solution) = 6.81 moles/Kg * 1Kg/1.85 L = 3.68 moles/L
M= 3.68 moles/L
Note:
- Would be wrong in this case to assume density of water = 1 Kg/L since the solution is heavily concentrated in glucose and therefore the density of water deviates from its pure value.
Presumptive tests, also known as preliminary tests or field tests, allow drugs to be quickly classified into a particular chemical group, but do not unequivocally identify the presence of a specific chemical compound.
