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2 years ago

Mr. Ragusa asks Hassan to make silver crystals from the following reaction.

Chemistry

1 answer:

SVEN [57.7K]2 years ago

6 0

Answer:

Percentage yield = 61.7%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated:

2AgNO₃ + Cu —> Cu(NO₃)₂ + 2Ag

Next, we shall determine the mass of AgNO₃ that reacted and the mass of Ag produced from the balanced equation. This is illustrated below:

Molar mass of AgNO₃ = 108 + 14 + (16×3)

= 108 + 14 + 48

= 170 g/mol

Mass of AgNO₃ from the balanced equation = 2 × 170 = 340 g

Molar mass of Ag = 108 g/mol

Mass of Ag from the balanced equation = 2 × 108 = 216 g

SUMMARY:

From the balanced equation above,

340 g of AgNO₃ reacted to produce 216 g of Ag.

Next, we shall determine the theoretical yield of Ag. This can be obtained as follow:

From the balanced equation above,

340 g of AgNO₃ reacted to produce 216 g of Ag.

Therefore, 51 g of AgNO₃ will react to produce = (51 × 216)/340 = 32.4 g of Ag.

Thus, the theoretical yield of Ag is 32.4 g.

Finally, we shall determine the percentage yield of Ag. This can be obtained as follow:

Actual yield = 20 g

Theoretical yield = 32.4 g

Percentage yield =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield = 20 / 32.4 × 100

Percentage yield = 61.7%

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A sample of an ionic compound that is often used as a dough conditioner is analyzed and found to contain 4.628 g of potassium, 9

Roman55 [17]

Answer:

The empirical formula of the compound is = $KBrO_3$

The name of the compound is potassium bromate.

Explanation:

Mass of potassium = 4.628 g

Moles of potassium = $\frac{4.628 g}{39 g/mol}=0.1187 mol$

Mass of bromine = 9.457 g

Moles of bromine = $\frac{9.457 g}{80 g/mol}=0.1182 mol$

Mass of oxygen = 5.681 g

Moles of oxygen = $\frac{5.681 g}{16 g/mol}0.3551$

For empirical formula of the compound, divide the least number of moles from all the moles of elements present in the compound:

Potassium :

$\frac{0.1187 mol}{0.1182 mol}=1.0$

Bromine;

$\frac{0.1182 mol}{0.1182 mol}=1.0$

Oxygen ;

$\frac{0.3551 mol}{0.1182 mol}=3.0$

The empirical formula of the compound is = $KBrO_3$

The name of the compound is potassium bromate.

5 0

3 years ago

How many liters of CO2 are in 4.76 moles? (at STP)

dem82 [27]

Answer: The volume of carbon dioxide gas at STP for given amount is 106.624 L

Explanation:

We are given:

Moles of carbon dioxide = 4.76 moles

At STP:

1 mole of a gas occupies a volume of 22.4 Liters

So, for 4.76 moles of carbon dioxide gas will occupy a volume of = $\frac{22.4L}{1mol\times 4.76mol=106.624L$

Hence, the volume of carbon dioxide gas at STP for given amount is 106.624 L

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3 years ago

Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

zysi [14]

Answer: The value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$N_2+2O_2\rightarrow 2NO_2$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

We are given:

Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = $\frac{121.77}{1}\times 1.90=231.36 J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

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3 years ago

What is the mass in milligrams of 4.30 moles of sodium? use significant figures?

alexandr1967 [171]

Answer is: mass of 4,30 moles of sodium is 98800 mg.
n(Na) = 4,30 mol.
m(Na) = ?
m(Na) = n(Na) · M(Na).
m(Na) = 4,30 mol · 23 g/mol.
m(Na) = 98,90 g.
m(Na) = 98,90 g · 1000 mg/1g.
m(Na) = 98900 mg.
n - amount of substance.
m - mass of substance.
M - molar mass of substance.

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3 years ago

Which part of the flower produces seeds? (992 points)

Nata [24]

Answer:

The pistil, B.

Explanation:

The seeds are formed from the ovule which is contained in the ovary; all in the base, the pistil.

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2 years ago