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5 days ago

compute the mass of CaSO4 that can be prepared by the reaction of 3.2900g of H2SO4 with 3.1660g of CaCO3

2 answers:

8 0

<span>4.3065 g To begin with, consult the atomic masses for each involved element. Atomic weight of Calcium = 40.078 Atomic weight of Carbon = 12.0107 Atomic weight of Hydrogen = 1.00794 Atomic weight of Oxygen = 15.999 Atomic weight of Sulfur = 32.065 Next, compute the molar masses of both reactants and the product. Molar mass H2SO4 = 2 * 1.00794 + 32.065 + 4 * 15.999 = 98.07688 g/mol Molar mass CaCO3 = 40.078 + 12.0107 + 3 * 15.999 = 100.0857 g/mol Molar mass CaSO4 = 40.078 + 32.065 + 4 * 15.999 = 136.139 g/mol The balanced equation for the reaction between H2SO4 and CaCO3 is: CaCO3 + H2SO4 ==> CaSO4 + H2O + CO2 Thus, 1 mole each of CaCO3 and H2SO4 is necessary to generate 1 mole of CaSO4. Let's check the amount of moles we have for CaCO3 and H2SO4. CaCO3: 3.1660 g / 100.0857 g/mol = 0.031632891 mol H2SO4: 3.2900 g / 98.07688 g/mol = 0.033545113 mol H2SO4 is in slight excess, therefore CaCO3 is the limiting reactant, suggesting we can expect 0.031632891 moles of product. To find the mass, multiply the number of moles by the molar mass calculated previously. 0.031632891 mol * 136.139 g/mol = 4.306470148 g Given that we have 5 significant figures from our data, we round the final result to 5 figures, yielding 4.3065 g</span>

KiRa [2.7K]5 days ago

7 0

Answer: A total of 4.3020 grams of calcium sulfate will be generated.

Explanation:

To determine the number of moles, we utilize the formula:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

For calcium carbonate:

Mass of calcium carbonate given = 3.1660 g

Molar mass of calcium carbonate = 100.0869 g/mol

When we insert these values into the above formula, we find:

$\text{Moles of calcium carbonate}=\frac{3.1660g}{100.0869g/mol}=0.0316mol$

For sulfuric acid:

Mass of sulfuric acid given = 3.2900 g

Molar mass of sulfuric acid = 98.079 g/mol

Inserting these values into the prior formula produces:

$\text{Moles of sulfuric acid}=\frac{3.2900g}{98.079g/mol}=0.0335mol$

In relation to the indicated reaction:

$CaCO_3+H_2SO_4\rightarrow CaSO_4+H_2CO_3$

Based on the stoichiometry of the reaction:

1 mole of calcium carbonate reacts with 1 mole of sulfuric acid

Therefore, 0.0316 moles of calcium carbonate will react with = $\frac{1}{1}\times 0.0316mol=0.0316mol$

Since the provided amount of sulfuric acid exceeds the necessary quantity, it serves as the excess reagent.

Calcium carbonate acts as the limiting reagent as it restricts the yield of the product.

Referring to the stoichiometry of the reaction:

1 mole of calcium carbonate creates 1 mole of calcium sulfate

Thus, 0.0316 moles of calcium carbonate will yield = $\frac{1}{1}\times 0.0316mol=0.0316mol$ of calcium sulfate.

To compute the weight of calcium sulfate, we apply equation 1:

Molar mass of calcium sulfate = 136.14 g/mol

Moles of calcium sulfate = 0.0316 mol

Substituting values into equation 1 gives us:

$0.0316mol=\frac{Mass of calcium sulfate}}{136.14g/mol}\\\\\text{Mass of calcium sulfate}=4.3020g$

Thus, 4.3020 grams of calcium sulfate will be produced.

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Respuesta:

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Explicación:

5 H2O2(aq) + 2 MnO4-(aq) + 6 H+(aq) → 2 Mn2+(aq) + 8 H2O(l) + 5 O2(g)

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Si se descompone esta reacción en sus mitades de reducción y oxidación

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H202 se convierte en H2O y O2

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18 days ago

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Answer:

The enthalpy of the second intermediate equation is altered by halving its value and changing the sign.

Explanation:

Let's examine both the first and second intermediate reactions alongside the overall equation concerning the examined process;

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Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ

Second reaction;

2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ

Thus, the overall reaction becomes;

CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH =?

According to Hess's law, which states that the total heat change in a reaction is equal to the sum of the heat changes for each step, we cannot simply sum the enthalpies for this overall reaction. Instead, we obtain the overall enthalpy by halving the second intermediate reaction's enthalpy and changing its sign before adding, as illustrated below;

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1 month ago

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21 day ago

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11 days ago

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