Can someone pls help me on this

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

The activation energy for the reaction NO2(g)+CO(g)⟶NO(g)+CO2(g) is Ea = 75 kJ/mol and the change in enthalpy for the reaction i

lapo4ka [179]

Answer:

= 100kJ

Explanation:

The reverse reaction's activation energy of a reaction is the activation energy of the forward reaction plus ΔH of the reaction:

Ea of forward reaction =75kJ

∆H = -175 kJ/mol

Ea of reverse reaction = 75 +(-175)

= 100kJ

Note that a reverse reaction is one which can proceed in both direction depending on the conditions.

6 0

3 years ago

A balloon containing N2 has 1.70 moles and occupies 3.80L. What will the volume be if the number of moles is increased to 2.60mo

leonid [27]

Answer:

5.81L

Explanation:

N1 = 1.70 moles

V1 = 3.80L

V2 = ?

N2 = 2.60 moles

Mole - volume relationship,

N1 / V1 = N2 / V2

V2 = (N2 × V1) / N1

V2 = (2.60 × 3.80) / 1.70

V2 = 9.88 / 1.70

V2 = 5.81 L

The volume of the gas is 5.81L

8 0

3 years ago

Classify each element. Note that another term for main group is representative, another term for semimetal is metalloid, and the

NikAS [45]

The question is incomplete, here is the complete question:

Classify each element. Note that another term for main group is representative, another term for semi-metal is metalloid, and the inner transition metals are also called the lanthanide and actinide series.

Hf, Am, In, Ta, As, Se, Rn

Answer:

Hafnium and tantalum are transition elements.

Americium is a inner transition element.

Indium, Selenium and Radon are main group elements.

Arsenic is a metalloid.

Explanation:

Main group elements are the elements which belong to s block and p block. They are also known as representative elements.

S-block elements are defined as the elements whose last electron enters s-sub shell. The general electronic configuration of these elements is $ns^{1-2}$

P-block elements are defined as the elements whose last electron enters p-sub shell. The general electronic configuration of these elements is $np^{1-6}$

Metalloids are defined as the elements which show intermediate properties between metals and non-metals. There are 7 metalloids in the periodic table. They are: Boron, Silicon, germanium, Arsenic, Antimony, Tellurium and Polonium.

Transition elements are known as d-block elements. D block elements are defined as the elements whose last electron enters d sub shell. The general electronic configuration of these elements is $[(n-1)d^{1-10}ns^{0-2}]$

Inner transition elements are known as (f block) elements. (F block) elements are defined as the elements whose last electron enters (f subshell). The general electronic configuration of these elements is $[(n-2)f^{1-14}(n-1)d^{0-1}ns^{2}]$ . They are also known as lanthanide and actinide series.

For the given elements:

Option 1: Hf

Hafnium is the 72nd element of the periodic table having electronic configuration of $[Xe]4f^{14}5d^26s^2$

As, the last electron is entering the d subshell, it is a transition element.

Option 2: Am

Americium is the 95th element of the periodic table having electronic configuration of $[Rn]5f^{7}6d^07s^2$

As, the last electron is entering the (f subshell), it is a inner transition element.

Option 3: In

Indium is the 49th element of the periodic table having electronic configuration of $[Kr]5s^25p^1$

As, the last electron is entering the p subshell, it is a main group element.

Option 4: Ta

Tantalum is the 73rd element of the periodic table having electronic configuration of $[Xe]4f^{14}5d^56s^2$

As, the last electron is entering the d subshell, it is a transition element.

Option 5: As

Arsenic is the 33rd element of the periodic table having electronic configuration of $[Ar]4s^24p^3$

As, the last electron is entering the p subshell, it is a main group element. It shows an intermediate property of metal and non-metal. Thus, it is a metalloid.

Option 6: Se

Selenium is the 34th element of the periodic table having electronic configuration of $[Ar]4s^24p^4$

As, the last electron is entering the p subshell, it is a main group element.

Option 7: Rn

Radon is the 86th element of the periodic table having electronic configuration of $[Xe]4f^{14}5d^{10}6s^26p^6$

As, the last electron is entering the p subshell, it is a main group element.

5 0

3 years ago

Which of the following is true for balancing equations?

algol [13]

Answer:

A.

Explanation:

An equation with the equal amount and proportion of atoms of each element on both sides of the reaction is commonly referred to as a balanced chemical equation.

The law of conservation of matter asserts that no observable and empirical change in the amount of matter occurs within a conventional chemical process. As a result, each element in the product would have the same equal amount or numbers of atoms as the reactants.

5 0

3 years ago