By the reaction of aluminium with 40 grams of metallic oxide and high temperature, 28 grams of the metal and 25.5 grams of aluminium oxide are produced. How many grams of aluminium are required for the reaction?
OR
The chemical equation for the reaction between sulphuric acid and sodium hydroxide is:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
How many grams of sodium hydroxide will be required to react completely with 4.9 grams of sulphuric acid?
(H=1, O=16, Na=23, S=32). [3]
Given
- Mass of metallic oxide = 40 grams
- Mass of metal produced = 28 grams
- Mass of aluminum oxide produced = 25.5 grams
Let’s assume the mass of aluminum required for the reaction is xx grams.
According to the law of conservation of mass:
Mass of reactants = Mass of products
⇒ 40 + x = 28 + 25.5
⇒ x = 53.5- 40 = 13.5 g
∴ 13.5 grams of aluminum are required for the reaction.
OR
Molar mass of H2SO4 = 2 + 32 + 4×16 = 98 g
Molar mass of NaOH = 23 + 16 + 1 = 40 g
From the balanced chemical equation:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
98 g 80 g
Amount of NaOH required to completely react with 98 g H2SO4 = 80 g
∴ Amount of NaOH required to completely react with 4.9 g H2SO4 = 80\over 98 × 4.9 = 4.0 g
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