Answer:
liquid bolling
Explanation:
because it just using heat.
Answer:
The mass of copper(II) sulfide formed is:
= 81.24 g
Explanation:
The Balanced chemical equation for this reaction is :

given mass= 54 g
Molar mass of Cu = 63.55 g/mol

Moles of Cu = 0.8497 mol
Given mass = 42 g
Molar mass of S = 32.06 g/mol

Moles of S = 1.31 mol
Limiting Reagent :<em> The reagent which is present in less amount and consumed in a reactio</em>n
<u><em>First find the limiting reagent :</em></u>

1 mol of Cu require = 1 mol of S
0.8497 mol of Cu should require = 1 x 0.8497 mol
= 0.8497 mol of S
S present in the reaction Medium = 1.31 mol
S Required = 0.8497 mol
S is present in excess and <u>Cu is limiting reagent</u>
<u>All Cu is consumed in the reaction</u>
Amount Cu will decide the amount of CuS formed

1 mole of Cu gives = 1 mole of Copper sulfide
0.8497 mol of Cu = 1 x 0.8497 mole of Copper sulfide
= 0.8497
Molar mass of CuS = 95.611 g/mol


Mass of CuS = 0.8497 x 95.611
= 81.24 g
Answer:
21.9 g C
Explanation:
Fe₂O₃ + 3 C ⇒ 2 Fe + 3 CO
This is your chemical equation.
To find the grams of C required, first convert grams of Fe₂O₃ to moles using the molar mass (159.69 g/mol).
(97.1 g)/(159.69 g/mol) = 0.608 mol Fe₂O₃
Then, use the mole ratio between Fe₂O₃ and C to convert moles of Fe₂O₃ to moles of C. You can find the mole ratio by looking at the chemical equation.
(0.608 mol Fe₂O₃) × (3 mol Fe₂O₃/1 mol Fe₂O₃) = 1.824 mol C
You can then convert moles of C to grams using the molar mass (12.01 g/mol).
(1.824 mol) × (12.01 g/mol) = 21.9 g