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3 years ago

Given 450.98 g of Cu(NO3)2, how many moles of Ag can be made? Provide your final answer rounded to two decimal places.

Chemistry

1 answer:

vivado [14]3 years ago

6 0

Answer:

4.82 moles of Ag.

Explanation:

We'll begin by calculating the number of mole in 450.98 g of Cu(NO₃)₂. This can be obtained as follow:

Molar mass of Cu(NO₃)₂ = 63.5 + 2[14 + (16×3)]

= 63.5 + 2[14 + 48]

= 63.5 + 2[62]

= 63.5 + 124

= 187.5 g/mol

Mass of Cu(NO₃)₂ = 450.98 g

Mole of Cu(NO₃)₂ =?

Mole = mass /Molar mass

Mole of Cu(NO₃)₂ = 450.98 / 187.5

Mole of Cu(NO₃)₂ = 2.41 moles

Next, we shall determine the number of mole of Cu needed to produce 450.98 g (i.e 2.41 moles) of Cu(NO₃)₂. This can be obtained as follow:

Cu + 2AgNO₃ —> Cu(NO₃)₂ + 2Ag

From the balanced equation above,

1 mole of Cu reacted to produce 1 mole of Cu(NO₃)₂.

Therefore, 2.41 moles of Cu will also react to produce 2.41 moles of Cu(NO₃)₂.

Thus, 2.41 moles of Cu is needed for the reaction.

Finally, we shall determine the number of mole of Ag produced from the reaction. This can be obtained as follow:

From the balanced equation above,

1 mole of Cu reacted to produce 2 moles of Ag.

Therefore, 2.41 moles of Cu will react to produce = 2× 2.41 = 4.82 moles of Ag.

Thus, 4.82 moles of Ag were obtained from the reaction.

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The correct answer is 5 moles of CO2 are produced.

Explanation:

The given reaction:

CH₄ (g) + 2O₂ (g) ⇔ CO₂ (g) + 2H₂O (g)

The given reaction is an illustration of a combustion reaction. Any reaction in which a substance is burnt in excess of oxygen to generate water and carbon dioxide is termed as a combustion reaction. From the given equation, it is clear that the moles of the formation of the products are in the ratio 1: 2, that is, if 10 moles of H₂O is produced, the production of 5 moles of CO₂ will be produced.

Let us multiply, the given equation with 5 we get,

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3 years ago

A calorimeter weighing 123.7g has a quantity of 20C water added to it. The combined mass of the calorimeter + cold water is 198.

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Answer:

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Explanation:

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Calculate the amount of heat released in the combustion of 10.5 grams of Al with 3 grams of O2 to form Al2O3(s) at 25°C and 1 at

ElenaW [278]

Answer : The amount of heat released in the combustion is, 209.5 kJ

Explanation :

First we have to calculate the moles of Al and $O_2$ .

$\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=\frac{10.5g}{27g/mole}=0.389moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{3g}{32g/mole}=0.188moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

$4Al+3O_2\rightarrow 2Al_2O_3$

From the balanced reaction we conclude that

As, 3 mole of $O_2$ react with 4 mole of $Al$

So, 0.188 moles of $O_2$ react with $\frac{4}{3}\times 0.188=0.251$ moles of $Al$

From this we conclude that, $Al$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Al_2O_3$