50cm3 of a mixture of methane and hydrogen were mixed with excess oxygen and exploded, the product after cooling to the original
1 answer:
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Answer:
The answer is:
(a)
(b) NaCl
(c) 0.211 g
Explanation:
Given:
The mass of NaCl,
= 0.0860 g
The molar mass of NaCl,
= 58.44 g/mol
The volume of ,
= 30.0 ml
or,
= 0.030 L
Molarity of ,
= 0.050 M
Moles of NaCl will be:
=
=
=
now,
Moles of will be:
(a)
The reaction is:
⇒
(b)
1 mole of NaCl react with,
= 1 mol of
0.0015 mol needs,
=
Available mol of NaCl < needed amount of NaCl
So,
The limiting reagent is "NaCl".
(c)
The precipitate formed,
=
=
Taking into account the reaction stoichiometry, 102 grams of Al₂O₃ are formed when 48 grams of O₂ react.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
4 Al + 3 O₂ → 2 Al₂O₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Al: 4 moles
- O₂: 3 moles
- Al₂O₃: 2 moles
The molar mass of the compounds is:
- Al: 27 g/mole
- O₂: 32 g/mole
- Al₂O₃: 102 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Al: 4 moles ×27 g/mole= 108 grams
- O₂: 3 moles ×32 g/mole= 96 grams
- Al₂O₃: 2 moles ×102 g/mole= 204 grams
The following rule of three can be applied: if by reaction stoichiometry 96 grams of O₂ form 204 grams of Al₂O₃, 48 grams of O₂ form how much mass of Al₂O₃?
<u><em>mass of Al₂O₃= 102 grams</em></u>
Finally, 102 grams of Al₂O₃ are formed when 48 grams of O₂ react.
Learn more about the reaction stoichiometry:
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