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3 years ago

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Assuming an efficiency of 34.90%, calculate the actual yield of magnesium nitrate formed from 139.6 g of magnesium and excess co

pper(II) nitrate.Mg+Cu(NO3)2⟶Mg(NO3)2+Cu

1 answer:

MakcuM [25]3 years ago

8 0

Answer:

300.44 g

Explanation:

The balanced equation for the reaction is given below:

Mg + Cu(NO3)2 —> Mg(NO3)2 + Cu

Next, we shall determine the mass of Mg that reacted and the mass of Mg(NO3)2 produced from the balanced equation.

This is illustrated below:

Molar mass of Mg = 24 g/mol

Mass of Mg from the balanced equation = 1 x 24 = 24 g

Molar mass of Mg(NO3)2 = 24 + 2[14 + (16x3)]

= 24 + 2[ 14 + 48]

= 24 + 124 = 148 g/mol

Mass of Mg(NO3)2 from the balanced equation =

1 x 148 = 148 g

From the balanced equation above,

24 g of Mg reacted to produce 148 g of Mg(NO3)2.

Next, we shall determine the theoretical yield of Mg(NO3)2.

This can be obtained as follow:

From the balanced equation above,

24 g of Mg reacted to produce 148 g of Mg(NO3)2.

Therefore, 139.6 g of Mg will react to = (139.6 x 148)/24 = 860.87 g of Mg(NO3)2

Therefore, the theoretical yield of Mg(NO3)2 is 860.87 g

Finally, we shall determine the actual yield of Mg(NO3)2 as follow:

Theoretical of Mg(NO3)2 = 860.87 g

Percentage yield = 34.90%

Actual yield of Mg(NO3)2 =?

Percentage yield = Actual yield /Theoretical yield x 100

34.90% = Actual yield /860.87

Cross multiply

Actual yield = 34.90% x 860.87

Actual yield = 34.9/100 x 860.87

Actual yield = 300.44 g

Therefore, the actual yield of Mg(NO3)2 is 300.44 g

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Now, we will calculate the partial pressure of heptane as follows.

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4 years ago

Use the following equation to answer the questions and please show all work.

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Answer:

a.36 g of water is produced.

b.64 g of $O_{2}$ is consumed.

Explanation:

The reaction is $2H_{2} + O_{2}$ ⇒ $2H_{2}O$

a.

Given,

Weight of $H_{2}$ reacted = 4g

Weight of 1 mole of $H_{2}$ = 2 $\times$ 1 = 2g

Therefore no. of moles of $H_{2}$ reacted = $\frac{4}{2}$ = 2 moles;

Also given,

Weight of $O_{2}$ reacted = 32 g

Weight of 1 mole of $O_{2}$ = 2 $\times$ 16 = 32 g

Therefore no. of moles of $O_{2}$ reacted = $\frac{32}{32}$ = 1

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As we took 2 moles of Hydrogen and 1 mole of Oxygen,

Directly,from the equation we can tell 2moles of water will be produced.

Therefore no. of moles of $H_{2} O$ produced = 2

Weight of 1 mole of water = $2\times 1+16$ = 18

Therefore weight of $H_{2}O$ produced = $2\times 18$ = 36gm

b.

Given ,

72 g of $H_{2}O$ is produced.

So,

no. of moles of $H_{2}O$ produced = $\frac{72}{18}$ = 4 moles

From equation For every 2 moles of water formed , 1 mole of oxygen must be required.

So for producing 4 moles of water,

No. of moles of Oxygen required = 2 moles.

Therefore weight of $O_{2}$ reacted = $2\times32$ = 64 g

Method 2:

Given,

8 g of $H_{2}$ has reacted.

So,

no. of moles of $H_{2}$ reacted = $\frac{8}{2}$ = 4 moles.

From equation , we know that For every 2 moles of $H_{2}$ reacted,1 mole of $O_{2}$ will react.

Therefore,

No. of moles of $O_{2}$ that reacts with 4 moles of $H_{2}$ = $2\times1$ = 2 moles

Therefore the weight of $O_{2}$ reacted = $2\times 32$ = 64 g

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