Answer:
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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:
As temperature increases the volume of given amount of gas increases while pressure and number of moles remain constant.
Explanation:
According to the charle's law,
The volume of given amount of gas is directly proportional to the temperature at constant pressure and number of moles of gas.
Mathematical expression:
V ∝ T
V = KT
V/T = K
When temperature changes from T₁ to T₂ and volume changes from V₁ to V₂.
V₁/T₁ = K V₂/T₂ = K
or
V₁/T₁ = V₂/T₂
Thus, the ratio of volume and temperature remain constant for constant amount of gas at constant pressure.
