How many grams can be prepared until equilibrium is attained? carbon disulfide is prepared by heating sulfur and charcoal. the c
1 answer:
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Answer:
The mass of oxygen that reacted is approximately 1.6 grams of oxygen
Explanation:
The given information are;
The mass of magnesium in the reaction = 2.43 grams
The final mass of the magnesium oxide = 4.12 grams
The chemical equation for the reaction is given as follows;
2Mg + O₂ → 2MgO
From which we see that two moles of magnesium, Mg, reacts with one mole of oxygen gas molecule, O₂, to produce two moles of magnesium oxide, MgO
The number of moles of magnesium present, , is given as follows;
By the given chemical equation, 2 moles of magnesium reacts with one mole of O₂, therefore;
1 mole of magnesium will react with 1/2 moles of oxygen which is 0.5 moles of oxygen also;
0.1 mole of magnesium will react with 0.05 moles of oxygen
The mass of one mole of oxygen = The molar mass of oxygen = 32 g/mol
The mass of oxygen in the reaction = The number of moles of oxygen × The molar mass of oxygen
The mass of oxygen in the reaction = 0.05 × 32 = 1.6 grams
Also from the molar mass of MgO which is 40.3044 g/mol, we have;
The mass fraction of oxygen = 16/40.3044 = 0.39698 ≈ 0.4
The mass of oxygen = 0.39698 × 4.12 = 1.636 g
Therefore, the mass of oxygen in the reaction is approximately 1.6 grams.
The energy needed to raise the temperature of water from 22.0ºC to 90.0ºC is c. 28.4 kJ.
<h3>What is specific heat?</h3>The amount of energy needed to raise the temperature of one gram of a substance by one degree Celsius.
By the formula
Q is the heat
m is the mass
c is the specific heat
Now, c = 4.184 J/g.K
The change in temperature is 22.0 ºC to 90.0 ºC
Putting the value in the equation
Thus, the energy needed to raise the temperature of water from 22.0ºC to 90.0ºC is 28.4 kJ
Learn more about specific heat
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Answer: The student should obtain <u>1.103 g of aluminum oxide </u>
Explanation:
- First we write down the equations that represent the aluminum hydroxide precipitation from the reaction between the aluminum nitrate and the sodium hydroxide:
Al(NO3)3 + 3NaOH → 3NaNO3 + Al(OH)3
Now, the equation that represents the decomposition of the hydroxide to aluminum oxide by heating it.
2Al(OH)3 → Al2O3 + 3H2O
- Second, we gather the information what we are going to use in our calculations.
Volumen of Al(NO3)3 = 40mL
Molar concentration of Al(NO3)3 = 0.541M
Molecular Weight Al2O3 = 101.96 g/mol
- Third, we start using the molar concentration of the aluminum nitrate and volume used to find out the total amount of moles that are reacting
then we use the molar coefficients from the equations to discover the amount of Al2O3 moles produced
finally, we use the molecular weight of the Al2O3 to calculate the final mass produced.