The bonds that are broken when water vaporizes are

For the following balanced equation: 3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l) a) How many moles of HNO3 will r

s344n2d4d5 [400]

Answer:

a) 26.67 moles HNO3

b) 0.33 moles NO

c) 0.40 moles NO is produced

d).157 moles Cu

e) 0.105 moles NO

f) 26.4 grams HNO3

g) Cu is in excess

h) 2.41 grams Cu remain

i) 2.37 grams NO

Explanation:

Step 1: Data given

Molar mass of Cu = 63.55 g/mol

Molar mass of HNO3 = 63.01 g/mol

Molar mass of Cu(NO3)2 = 187.56 g/mol

Molar mass of NO = 30.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation

3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l)

a) How many moles of HNO3 will react with 10 moles of Cu?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 10 moles Cu we need 8/3 *10 = 26.67 moles HNO3

b) How many moles of NO will form if 0.50 moles of Cu reacts?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.50 moles Cu we'll have 2/3 *0.50 = 0.33 moles NO

c) If 0.80 moles of H2O forms, how much NO must also form?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

If 0.80 moles H2O is produced, 0.80/2 = 0.40 moles NO is produced

d) How many moles of Cu are in 10.0 grams of Cu?

Moles Cu = 10.0 grams / 63.55 g/mol = 0.157 moles

In 10.0 grams Cu we have 0.157 moles Cu

e) If 10.0 g of Cu reacts, how many moles of NO will form?

10.0 grams Cu = 0.157 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.157 moles Cu we'll have 2/3 * 0.157 = 0.105 moles NO

f) If 10.0 g of Cu reacts, how many grams of HNO3 are required?

10.0 grams Cu = 0.157 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.157 moles Cu we'll need 0.419 moles HNO3

This is 0.419 moles * 63.01 g/mol = 26.4 grams HNO3

g) If 10.0 g of Cu and 20.0 g of HNO3 are put together in a reaction vessel, which one will be in excess?

Moles Cu = 0.157 moles

Moles HNO3 = 20.0 grams / 63.01 g/mo = 0.317 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

The limiting reactant is HNO3. It will completely be consumed (0.317 moles). Cu is in excess. There will react 3/8 * 0.317 = 0.119 moles Cu

There will remain 0.157 - 0.119 = 0.038 moles

h) How many grams of the excess substance will be left over?

There will react 3/8 * 0.317 = 0.119 moles Cu

There will remain 0.157 - 0.119 = 0.038 moles

This is 0.038 moles * 63.55 g/mol = 2.41 grams

i) How many grams of NO will form in the reaction described in part g?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.317 moles HNO3 we'll have 0.317/4 = 0.0793 moles NO

This is 0.079 mol * 30.01 g/mol = 2.37 grams NO

3 0

3 years ago

You have 4.5 x 10^24 particles of C3H8. How many grams are present?

Citrus2011 [14]

Answer:

33 g.

Explanation:

Hello there!

In this case, for these particle-mole-mass relationships problems, it is necessary for us to recall the following equivalence statement, based off the molar mass of the involved compound, C3H8, one mole of particles and the Avogadro's number:

$1mol=44.11g=6.022x10^{23}molecules$

In such a way, we can set up the following expression for the calculation of the mass in the given particles of propane:

$4.5x10^{23}molecules*\frac{1mol}{6.022x10^{23}molecules} *\frac{44.11g}{1mol}\\\\33g$

Best regards!

3 0

3 years ago

Please explain how we judge the brightness of stars.

11111nata11111 [884]

Astronomers can measure it by way of luminosity which is the power of a star or the amount of energy (light) the star admits from its surface. they also measure the brightness of a start as if it were to appear 32.6 light years from Earth

6 0

3 years ago

If 8.50 g of phosphorus reacts with hydrogen gas at 2.00 atm in a 10.0-L container at 298 K, calculate the moles of PH3 produced

ahrayia [7]

Answer:

The moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.

Explanation:

Phosphorus reacts with H₂ according to the balanced equation:

P₄ (s) + 6 H₂ (g) ⇒ 4 PH₃ (g)

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

P₄: 1 mole
H₂: 6 moles
PH₃:4 moles

Being the molar mass of the compounds:

P₄: 124 g/mole
H₂: 2 g/mole
PH₃: 34 g/mole

The following mass amounts of each compound participate in the reaction:

P₄: 1 mole* 124 g/mole= 124 g
H₂: 6 mole* 2 g/mole= 12 g
PH₃: 4 moles* 34 g/mole= 136 g

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case you know:

P= 2 atm
V= 10 L
n= ?
R= 0.082 $\frac{atm*L}{mol*K}$
T= 298 K

Replacing:

2 atm*10 L= n*0.082 $\frac{atm*L}{mol*K}$ *298 K

and solving you get:

$n=\frac{2 atm*10 L}{0.082\frac{atm*L}{mol*K}*298 K }$

n=0.818 moles

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, you can use a simple rule of three as follows: if 6 moles of H₂ react with 124 g of P₄, 0.818 moles of H₂ with how much mass of P₄ will it react?

$mass of P_{4}=\frac{0.818 moles of H_{2}*124 grams of P_{4}}{6 moles of H_{2}}$

mass of P₄= 16.90 grams

But 16.90 grams of P₄ are not available, 8.50 grams are available. Since you have less mass than you need to react with 0.818 moles of H₂, phosphorus P₄ will be the limiting reagent.

Then you can apply the following rules of three:

If 124 grams of P₄ produce 4 moles of PH₃, 8.50 grams of P₄, how many moles do they produce?

$moles of PH_{3} =\frac{8.5 grams of P_{4}*4 moles of PH_{3} }{124grams of P_{4}}$

moles of PH₃=0.2742

If 124 grams of P₄ react with 6 moles of H₂, 8.50 grams of P₄ with how many moles of H₂ do they react?

$moles of H_{2} =\frac{8.5 grams of P_{4}*6 moles of H_{2} }{124grams of P_{4}}$

moles of H₂= 0.4113

If you have 0.818 moles of H₂, the number of moles of gas H₂ present at the end of the reaction is calculated as:

0.818 - 0.4113= 0.4067

Then the total number of moles of gas present at the end of the reaction will be the sum of the moles of PH₃ gas and H₂ gas that did not react:

0.2742 + 0.4067= 0.6809

Finally, the moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.

5 0

3 years ago

What negative impact did chemistry have on society through the production and use of chlorofluorocarbons (CFCs)?

andrey2020 [161]

The answer to your question is the 3rd option

6 0

3 years ago

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