A 1.268 g sample of a metal carbonate, MCO₃, was treated 100.00 mL of 0.1083 M H₂SO₄, yielding CO₂ gas and an aqueous solution o
f the metal sulfate. The solution was boiled to remove all of the dissolved CO₂ and then was titrated with 0.1241 M NaOH. A 71.02 mL volume of the NaOH solution was required to neutralize the excess H₂SO₄.a) Write the balanced chemical equation for this reaction.
b) What is the identity of the metal?
c) How many grams of CO₂ gas were produced?
b) What is the identity of the metal?
c) How many grams of CO₂ gas were produced?
1 answer:
Answer:
MCO3 is BaCO3
The mass of CO2 produced is 0.28g of CO2
Explanation:
The first step in solving the question is to put down the balanced reaction equations as shown in the image attached. Secondly, we obtain the relative number of moles acid and base as mentioned in the question. The balanced neutralization reaction equation is used to obtain the number of moles of excess acid involved in the neutralization reaction.
This is then subtracted from the total number of moles acid to give the number of moles of acid that reacted with MCO3. From here, the molar mass of MCO3 and identity of M can be found. Hence the mass of CO2 produced is calculated as shown.
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Answer:
1 gramo de metano aporta 50.125 kilojoules.
1 gramo de metano aporta 48.246 kilojoules.
Explanation:
La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo (), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto (
), en kilojoules por mol, dividido por su masa molar (
), en gramos por mol:
(1)
A continuación, analizamos cada caso:
Metano
1 gramo de metano aporta 50.125 kilojoules.
Octano
1 gramo de metano aporta 48.246 kilojoules.