Answer:
Explanation:
First, we find in the tables the ΔH of formation of each compound. As you can see in the (image 1)
Then we solve the ecuation for ΔH°reaction
ΔH°reaction=∑ΔH°f(products)−∑ΔH°f(Reactants)
ΔH°reaction= (-2* 393.5 - 2*285.8) - (52.4 + 0) kJ/mol
ΔH°reaction = -1.41 *10^3 kJ/mol
Magnesium
A contains 38.5 g of tin for each 12.3 g of fluorine:
<span>mole ratio: </span>
<span>(38.5 g)/(118.71 g/mol):(12.3 g)/(18.998 g/mol) = 0.324:0.647 = 1:2 ⇒ SnF₂ </span>
<span>B contains 56.5 g of tin for each 36.2 g of fluorine: </span>
<span>mole ratio: </span>
<span>(56.5 g)/(118.71 g/mol):(36.2 g)/(18.998 g/mol) = 0.476:1.905 = 1:4 ⇒ SnF₄
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2ZnS(s)+3O2(g) -> 2Zns(s) + 2SO3(g)
the above given equation is unbalanced as it contains 4 moles of sulphur in the output but in the input there are only two aoms of sulphur so to balance the equation we will write the equation as given under
balanced equation is
2ZnS(s)+3O2(g) -> 2Zn(s) + 2SO3(g)
In words:
When 2 moles of solid zinc sulfide reacts with 3 moles of oxygen gas gives 2 moles of solid zinc and 2 moles of sulphur trioxide gas.
Answer:
"2.48 mole" of H₂ are formed. A further explanation is provided below.
Explanation:
The given values are:
Mole of Al,
= 3.22 mole
Mole of HBr,
= 4.96 mole
Now,
(a)
The number of mole of H₂ are:
⇒ 
or,
⇒ 
⇒ 
⇒ 
(b)
The limiting reactant is:
= 
(c)
The excess reactant is:
= 
