Answer:
The equation for the reaction of one sodium bicarbonate ( NaHCO3 ) molecule with one citric acid (C6H8O7) molecule is the following:
Sodium Bicarbonate + Citric Acid ⇒ Water + Carbon Dioxide + Sodium Citrate
NaHCO3 + C6H8O7 ⇒ 3 CO2 + 3 H2O + Na3C6H5O7
Explanation:
The reaction is in balance, that is, the whole H2CO3 is not finished, but a little bit of this acid is left in the solution. Therefore, when sodium bicarbonate is added to the solution with citric acid, sodium citrate salt (C6H5O7Na3) and carbonic acid (H2CO3) are formed, which is rapidly broken down into water (H2O) and carbonic oxide (CO2).
C6H8O7 + NaHCO3 ⇒ C6H5O7Na3 + 3 H2CO3
C6H5O7Na3 + 3 H2CO3 ⇔ C6H5O7Na3 + 3 H2O + 3 CO2
The balanced chemical equation for the above reaction is as follows ;
Mg + 2HCl —> MgCl2 + H2
The stoichiometry of Mg to HCl is 1:2
This means that 1 mol of Mg reacts with 2 mol of HCl
Equal amounts of both Mg and HCl have been added. One reagent is the limiting reactant and other reactant is in excess.
Limiting reactant is the reagent that is fully used up in the reaction and the amount of Product formed depends on the amount of limiting reactant present.
In this reaction if Mg is the limiting reactant, 4.40 moles of Mg should react with 4.40x2 -8.80 moles of HCl.
But only 4.40 moles of HCl present therefore HCl is the limiting reactant that reacts with 4.40/2 = 2.20 moles of Mg
Stoichiometry of HCl to MgCl2 is 2:1
Since HCl moles reacted -4.40 mol
Then MgCl2 moles formed are 4.40/2 = 2.20 mol of MgCl2
The problem talks about two questions and these are:
1. Metals are very good conductors of electricity and heat. Directing heat is easier. So let Marie heat the beads and also have heat another substance, for instance, water. If the beads heat quicker, then they are metals. Another test to conduct is called flame test. This test should give you a colored flame (blue/white for lead) the metal is lead if the reaction is: 2PbO+C ==> 2Pb +CO2
2. The beads are possibly to be lead since Ferrous(lead) oxide + carbon = carbon dioxide + lead
Answer:
334J/g
Explanation:
Data obtained from the question include:
Mass (m) = 1g
Specific heat of Fusion (Hf) = 334 J/g
Heat (Q) =?
Using the equation Q = m·Hf, we can obtain the heat released as follow:
Q = m·Hf
Q = 1 x 334
Q = 334J
Therefore, the amount of heat released is 334J
Answer:
58.6 % by mass of Na₂CO₃
Explanation:
This is the reaction:
Na₂CO₃ + MgCO₃ + 4HCl → MgCl₂ + 2NaCl + 2CO₂ + 2H₂O
Let's find out the moles of CO₂ produced, by the Ideal Gases Law
1.24 atm . 1.67 L = n . 0.082 . 299K
(1.24 atm . 1.67 L / 0.082 . 299K) = n
0.0844 moles = n
Ratio is 2:1, so 2 moles of dioxide were produced by 1 mol of sodium carbonate. Let's make a rule of three:
2 moles of CO₂ were produced by 1 mol of Na₂CO₃
Then, 0.0844 moles of Co₂ would beeen produced by (0.0844 .1)/2 = 0.0422 moles of Na₂CO₃.
Let's convert this moles into mass (mol . molar mass)
0.0422 mol . 106 g/mol = 4.47 g
Finally we can know the mass percent of sodium carbonate in the mixture
(Mass of compound /Total mass) . 100 → (4.47 g / 7.63g) . 100 = 58.6 %
