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Snezhnost [94]
3 years ago
7

Calcium oxide reacts with water to produce calcium hydroxide and heat as shown by the equation. CaO(s) + H2O(l) → Ca(OH)2(aq) +

heat When 33 g of CaO and 10 g of H2O react, how many grams of calcium hydroxide would you expect to be produced? Explain your answer.
Chemistry
1 answer:
GuDViN [60]3 years ago
3 0

Answer:

41.11 g of Ca(OH)2.

Explanation:

The balanced equation for the reaction is given below:

CaO(s) + H2O(l) → Ca(OH)2(aq)

Next, we shall determine the masses of CaO and H2O that reacted and the mass of Ca(OH)2 produced from the balanced equation. This can be obtained as follow:

Molar mass of CaO = 40 + 16 = 56 g/mol

Mass of CaO from the balanced equation = 1 × 56 = 56 g

Molar mass of H2O = (2×1) + 16 = 2 + 16 = 18 g/mol

Mass of H2O from the balanced equation = 1 × 18 = 18 g

Molar mass of Ca(OH)2 = 40 + 2(16 + 1)

= 40 + 2(17)

= 40 + 34

= 74 g/mol

Mass of Ca(OH)2 from the balanced equation = 1 × 74 = 74 g

From the balanced equation above,

56 g of CaO reacted with 18 g of H2O to produce 74 g of Ca(OH)2.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

56 g of CaO reacted with 18 g of H2O.

Therefore, 33 g of CaO will react with = (33 × 18)/56 = 10.6 g of H2O.

From the calculation made above, we can see that it will take a higher amount (i.e 10.6 g) of H2O than was given (i.e 10 g) to react completely with 33 g of CaO. Therefore, H2O is the limiting reactant.

Finally, we shall determine the mass of Ca(OH)2 produced from the reaction.

In this case, the limiting reactant will be use because it will give the maximum yield of Ca(OH)2.

The limiting is H2O and the mass of Ca(OH)2 produced can be obtained as follow:

From the balanced equation above,

18 g of H2O reacted to produce 74 g of Ca(OH)2.

Therefore, 10 g of H2O will react to produce = (10 × 74)/18 = 41.11 g of Ca(OH)2.

Thus, 41.11 g of Ca(OH)2 were obtained from the reaction.

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g a solution is made by mixing 500.0 mL of 0.037980.03798 M Na2sO4 Na2sO4 with 500.0 mL of 0.034280.03428 M NaOH NaOH . Complete
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Answer:

The concentration of the sodium and arsenate ions at the end of the reaction in the final solution

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Explanation:

Complete Question

A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)

Concentration in mol/L = (Number of moles) ÷ (Volume in L)

Number of moles = (Concentration in mol/L) × (Number of moles)

For Na₂HAsO₄

Concentration in mol/L = 0.03798 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03798 × 0.5 = 0.01899 mole

For NaOH

Concentration in mol/L = 0.03428 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03428 × 0.5 = 0.01714 mole

Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

0.01899        0.01714        0           0 (At time t=0)

(0.01899 - 0.1714) | 0 → 0.01714    0.01714 (end)

0.00185  | 0 → 0.01714    0.01714 (end)  

Hence, at the end of the reaction, the following compounds have the following number of moles

Na₂HAsO₄ = 0.00185 mole

This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction

NaOH = 0 mole

Na₃AsO₄ = 0.01714 moles

This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction

H₂O = 0.01714 moles

So, at the end of the reaction

Na⁺ has 0.0037 + 0.05142 = 0.05512 mole

(HAsO₄)²⁻ has 0.00185 mole

(AsO₄)³⁻ has 0.01714 mole.

And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L

Hence, the concentration of the sodium and arsenate ions at the end of the reaction is

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Hope this Helps!!!

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