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postnew [5]
2 years ago
12

PLEASE HELP! Find out whether the chemical equation is balanced or unbalanced.

Chemistry
1 answer:
nordsb [41]2 years ago
4 0

One chemical reaction is called the Haber process, a method for preparing ammonia by reacting nitrogen gas with hydrogen gas:


This equation shows you what happens in the reaction, but it doesn’t show you how much of each element you need to produce the ammonia. To find out how much of each element you need, you have to balance the equation — make sure that the number of atoms on the left side of the equation equals the number of atoms on the right.


You know the reactants and the product for this reaction, and you can’t change them. You can’t change the compounds, and you can’t change the subscripts, because that would change the compounds.


So the only thing you can do to balance the equation is add coefficients, whole numbers in front of the compounds or elements in the equation. Coefficients tell you how many atoms or molecules you have.


For example, if you write the following, it means you have two water molecules:


Each water molecule is composed of two hydrogen atoms and one oxygen atom. So with two water molecules (represented above), you have a total of 4 hydrogen atoms and 2 oxygen atoms.


You can balance equations by using a method called balancing by inspection. You take each atom in turn and balance it by adding appropriate coefficients to one side or the other.


With that in mind, take another look at the equation for preparing ammonia: HOPE THIS HELPS



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7.00 of Compound x with molecular formula C3H4 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25c.
beks73 [17]

Answer:

\Delta H_{f,C_3H_4}=276.8kJ/mol

Explanation:

Hello!

In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

\Delta H_{rxn} =- m_wC_w\Delta T

We plug in the mass of water, temperature change and specific heat to obtain:

\Delta H_{rxn} =- (35000g)(4.184\frac{J}{g\°C} )(2.316\°C)\\\\\Delta H_{rxn}=-339.16kJ

Now, this enthalpy of reaction corresponds to the combustion of propyne:

C_3H_4+4O_2\rightarrow 3CO_2+2H_2O

Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:

\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}

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:

\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol

Now, we solve for the enthalpy of formation of C3H4 as shown below:

\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}

So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):

\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol

Best regards!

7 0
2 years ago
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