What volume of 0.117 M HCl is needed to neutralize 28.67ml of 0.137 m KOH?

1 answer:

3 0

To determine the volume of the acid needed in the reaction, we first have to know the reaction involved in the process. Since it is a neutralization reaction, then it should produce a salt and water. The reaction would be:

HCl + KOH = KCl + H2O

We use this reaction to relate the substances. We calculate as follows:

0.137 M KOH (.02867 L solution ) = 3.92779x10^-3 mol KOH
3.92779x10^-3 mol KOH ( 1 mol HCl / 1 mol KOH ) = 3.92779x10^-3 mol HCl

Volume of HCl = 3.92779x10^-3 mol HCl / 0.117 M HCl
= 0.03357 L HCl or 33.57 mL HCl needed

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15. Ammonium nitrate, NH4NO3, and aluminum powder react explosively producing nitrogen gas, water vapor and aluminum oxide. Writ

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Answer:

$2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)$

$Entalpy=-2861.9~KJ$

Explanation:

In this case, we have to start with the reagents:

$Al~+~NH_4NO_3$

The compounds given by the problem are:

-) Nitrogen gas = $N_2$

-) Water vapor = $H_2O$

-) Aluminum oxide = $Al_2O_3$

Now, we can put the products in the reaction:

$Al_(_S_) ~+~NH_4NO_3_(_aq_) ->N_2_(_g_) ~+~H_2O_(_g_) ~+~Al_2O_3_(_S_)$

When we balance the reaction we will obtain:

$2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)$

Now, for the enthalpy change, we have to find the standard enthalpy values:

$Al_(_S_)=0~KJ/mol$

$NH_4NO_3_(_a_q_)=-132.0~KJ/mol$

$N_2_(_g_)=0~KJ/mol$

$H_2O_(_g_)=~-~241.8~KJ/mol$

$Al_2O_3_(_S_)=~-~1675.7~KJ/mol$

With this in mind, if we multiply the number of moles (in the balanced reaction) by the standard enthalpy value, we can calculate the energy of the reagents:

$(0*2)~+~(-132*3)=~-396~KJ$