When aluminum is placed in concentrated hydrochloric acid, hydrogen gas is produced. 2 Al ( s ) + 6 HCl ( aq ) ⟶ 2AlCl3 ( aq ) +
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Answer: The volume of hydrogen gas produced will be, 12.4 L
Explanation : Given,
Mass of = 54.219 g
Number of atoms of =
Molar mass of = 98 g/mol
First we have to calculate the moles of and
.
and,
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
From the balanced reaction we conclude that
As, 3 mole of react with 2 mole of
So, 0.553 moles of react with
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of
From the reaction, we conclude that
As, 3 mole of react to give 3 mole of
So, 0.553 mole of react to give 0.553 mole of
Now we have to calculate the volume of gas at STP.
As we know that, 1 mole of substance occupies 22.4 L volume of gas.
As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas
So, 0.553 mole of hydrogen gas occupies volume of hydrogen gas
Therefore, the volume of hydrogen gas produced will be, 12.4 L
Answer:
Explanation:
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In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:
We plug in the mass of water, temperature change and specific heat to obtain:
Now, this enthalpy of reaction corresponds to the combustion of propyne:
Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:
However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:
Now, we solve for the enthalpy of formation of C3H4 as shown below:
So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):
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