How many grams of Al would be needed to displace all of the copper ions in 71.8 mL of 0.610 M CuSO4?
1 answer:
The amount of Al that would be needed will be 0.79 grams
<h3>Stoichiometric calculations</h3>From the equation of the reaction below:
The mole ratio of Al to is 2:3.
Mole of 71.8 mL. 0.610 M = 0.610 x 71.8/1000 = 0.0438 moles
Equivalent mole of Al = 2/3 x 0.0438 = 0.029 moles
Mass of o.o29 moles Al = 0.029 x 26.98 = 0.79 grams
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Answer:
a. Δ
b. 320.76° C
Explanation:
a.)
we can solve this type of question (i.e calculate Δ , for the gas-phase reaction ) using the Hess's Law.
Δ =
Given from the question, the table below shows the corresponding Δ for each compound.
Compound
Liquid EO -77.4
-74.9
-110.5
If we incorporate our data into the above previous equation; we have:
Δ = (-110.5 kJ/mol + (-74.9 kJ/mol) ) - (-77.4 kJ/mol)
=
b.)
We are to find the final temperature if the average specific heat capacity of the products is 2.5 J/g°C
Given that:
the specific heat capacity (c) = 2.5 J/g°C
= 93.0°C &
the enthalpy of vaporization (Δ) = 569.4 J/g
If, we recall; we will remember that; Specific Heat Capacity is the amount of heat needed to raise the temperature of one gram of a substance by one kelvin.
∴ the specific heat capacity (c) is given as =
Let's not forget as well, that Δ =
If we substitute Δ for
in the above equation, we have;
specific heat capacity (c) =
Making () the subject of the formula; we have:
=
= 227.76°C +93.0°C
= 320.76°C
∴ we can thereby conclude that the final temperature = 320.76°C