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Reika [66]
3 years ago
8

21 pts. Sodium reacts with oxygen to produce sodium oxide as described by the balanced equation below. If 54.1g of sodium reacts

with excess oxygen gas to produce 61.8g of sodium oxide, what is the percent yield? Show all work. (hint: be sure to calculate theoretical yield first) 4Na + O2 --> 2Na2O
Chemistry
1 answer:
Aleksandr [31]3 years ago
8 0
First, calculate the amount fo Na2O that should be produced with the given amount of sodium using proper dimensional analysis and  the balanced chemical reaction,

   theoretical amount of Na2O = (54.1 g of Na)(1 mol Na/23 g Na)(2 mos Na2O/4 mols Na)(62 g Na2O/ 1mol Na2O)
       
      theoretical amount of Na2O = 72.9 g of Na2O


Calculation of percent yield,

                % yield = (61.8 g Na2O) / (72.9 g of Na2O) x 100%
                 % yield = 84.75%

Answer: 84.75%
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What temperature will the water reach when 10.1 g CaO is dropped into a coffee cup containing 157 g H2O at 18.0°C if the followi
Zepler [3.9K]

Answer:

Final temperature attained by water = 34.6°C

Explanation:

The reaction of CaO and H₂O is an <em>exothermic reaction</em>. The equation of reaction is given below:

CaO + H₂O ----> Ca(OH)₂

The quantity of heat given off, ΔH°rxn = 64.8KJ/mol = 64800J/mol

Number of moles of CaO = mass/molar mass, where molar mass of Ca0 = 56g/mol, mass of CaO = 10.1g

Number of moles of CaO = 10.1g/56g/mol =0.179moles

Quantity of heat given off by 0.179 moles = 64800 *0.179 = 11599.2J/mol

Using the formula, <em>Quantity of heat, q = mass * specific heat capacity * temperature rise.</em>

mass of mixture = (10.1 + 157)g = 167.1g, Initial temperature = 18.0°C

Final temperature(T₂) - Initial temperature(T₁) = Temperature rise

11599.2J/mol = 167.1g * 4.18J/g·°C * ( T₂ - 18.0°C)

11599.2 = 698.478T₂ - 12572.604

11599.2 + 12572.604 = 698.478T₂

698.478T₂ = 24171.804

T₂ = 34.6°C

Therefore, final temperature attained by water = 34.6°C

6 0
3 years ago
Calculate the radius of tantalum (Ta) atom, given that Ta has a BCC crystal structure, a density of 16.6 g/cm, and an atomic wei
Ivahew [28]

Answer:

The radius of tantalum (Ta) atom is R = 1.43 \times 10^{-8} \:cm = 0.143 \:nm

Explanation:

From the Body-centered cubic (BBC) crystal structure we know that a unit cell length <em>a </em>and atomic radius <em>R </em>are related through

a=\frac{4R}{\sqrt{3} }

So the volume of the unit cell V_{c} is

V_{c}= a^3=(\frac{4R}{\sqrt{3} } )^3=\frac{64\sqrt{3}R^3}{9}

We can compute the theoretical density ρ through the following relationship

\rho=\frac{nA}{V_{c}N_{a}}

where

n = number of atoms associated with each unit cell

A = atomic weight

V_{c} = volume of the unit cell

N_{a} =  Avogadro’s number (6.023 \times 10^{23} atoms/mol)

From the information given:

A = 180.9 g/mol

ρ = 16.6 g/cm^3

Since the crystal structure is BCC, n, the number of atoms per unit cell, is 2.

We can use the theoretical density ρ to find the radio <em>R</em> as follows:

\rho=\frac{nA}{V_{c}N_{a}}\\\rho=\frac{nA}{(\frac{64\sqrt{3}R^3}{9})N_{a}}

Solving for <em>R</em>

\rho=\frac{nA}{(\frac{64\sqrt{3}R^3}{9})N_{a}}\\\frac{64\sqrt{3}R^3}{9}=\frac{nA}{\rho N_{a}}\\R^3=\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}} \\R=\sqrt[3]{\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}

Substitution for the various parameters into above equation yields

R=\sqrt[3]{\frac{2\cdot 180.9}{16.6\cdot 6.023 \times 10^{23}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}\\R = 1.43 \times 10^{-8} \:cm = 0.143 \:nm

7 0
3 years ago
which statement best describes the two tails of a comet? heating by the sun, the comets rotation, gravitational effects of nearb
Firdavs [7]

Answer: the comets rotation

Explanation:

5 0
3 years ago
What does electron have to do with color ? In one sentence
Goryan [66]

Electrons absorb energy, as they absorb energy they go from ground state to excited state and to return to ground state electrons release energy in the form of photons producing that color.

4 0
3 years ago
Need help on some homework questions I’m stuck on about molarity and dilutions:
Flura [38]

  question 1

The  moles of C12H22O11  that  are in  7.5 L  of a  5.8 M C12H22O11 solution  is    43.5  moles


 <u><em>calculation</em></u>

<u><em> </em></u>moles = molarity  x volume

volume = 7.5 L

molarity = 5.8 M or  5.8 mol/l

moles is therefore = 7.5 l x 5.8 mol/l = 43.5 moles


question 2

% percent  =( mas of the solute/ mass of the solution)  x 100

mass of solute= ?

mass of the solution   = 375 g

% mass =  85 %  = 85/100

let the mass of solute be represented by Y

therefore  y/375  = 85/100

y=(375 x 85)/100 = 318.75 g


moles =  mass/molar mass

= 318.75 g/ 42.4= 7.52  moles


Question  3

 The grams of copper  metal is  857.25 g

 calculate the  moles  HCl

=  molarity x volume  = 4.50 L x 6 .00 mol/l= 27 moles

  By use of mole ratio of Cu: HCl which is  1:2 the moles of Cu = 27/2 =13.5 moles

mass  = moles  x molar mass

= 13.5  moles x 63.5  g/mol = 857.25 g


question 4

Complete description of how to make  3.5 L of 2.0 M NaCl solution  from a stock of 6.0 M NaCl  is   that measure  1.17 L  of 6.0 M NaCl, Dissolve it  to 3.5 L to make 2.0 M NaCl


<u><em>calculation</em></u>

M1V1 = M2V2

m1=  2.0 M

V1=3.5 L

M2= 6.0 M

V2=?

v2  = M1V1/M2

= (3.5 L x  2.0 M)/ 6.0 M= 1.17 L

explanation : measure  1.17  L of 6.0M NaCl  and then dissolve  it to 3.5 L  to make 2.0 M NaCl


question 5

The number of grams  of stock  solution that is in 90.0 %  is HCl by mass  that would be  needed  to  make 175 g is 77.78 grams


calculation

This  is calculated using M1C1= M2C2 formula

M1  = ?g

C1  =90.0%

M1= 175 g

C1= 40.0 %

make M1 the subject of the  formula

M1= M2C2/C1

M1 is  therefore =( 175 g x 40) / 90 = 77.78 g



Question  6

The grams  of silver metal is 127 g

% mass =  mass of solute/ mass of solution

% mass  = 80% = 80/100

mass of solute (AgNO3)= ?

mass  of solution = 250 g

let the mass of solute be  represented by Y

therefore  Y/ 250 = 80/100

Y= (250 x 80) /100 =200 g  of  AgNO3

moles = mass/molar mass

moles of AgNO3 =  200 g/ 169.87 g/mol =1.178  moles

The mole ratio of AgNO3: Ag is 2:2=1:1  therefore the moles of Ag= 1.178 moles


mass=  moles  x molar mass

= 1.178  moles x  107.87 g/mol =127.07 g

6 0
3 years ago
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