What is the maximum number of grams of ph3 that can be formed when 6.2 g of phosphorus reacts with 4.0 g of hydrogen to form ph3? (do not key units into your answer)

p4(g) + 6h2(g) → 4ph3(g)

The maximum mass of phosphane  

(

PH

3

)

that can be produced under the conditions stated in the question is  

6.8 g

.

Explanation:

Start with a balanced equation.

P

4

+

6H

2

→

4PH

3

This is a limiting reactant problem. The reactant that produces the least  

PH

3

determines the maximum number of grams of  

PH

3

The process will go as follows:

given mass P

4

→

mole P

4

→

mol PH

3

→

mass PH

3

and

given mass H

2

→

mole H

2

→

mol PH

3

→

mass PH

3

The molar masses of each substance is needed. Molar mass is the mass of one mole of an element, molecule or ionic compound in g/mol.

Molar Masses

Multiply the subscript for each element by its atomic weight on the periodic table in g/mol. If there is more than one element, add the molar masses together.

P

4

:

(

4

×

30.974

g/mol P

)

=

123.896 g/mol P

4

H

2

:

(

2

×

1.008

g/mol H

)

=

2.016 g/mol H

2

PH

3

:

(

1

×

30.974

g/mol P

)

+

(

3

×

1.008

g/mol H

)

=

33.998 g/mol PH

3

Now you need to determine the mass of phosphane,  

PH

3

, produced individually by  

6.2 g P

4

and by  

4.0 g H

2

. I'm going to start with  

P

4

since it is first in the equation.

Moles of Phosphorus

Multiply the given mass of  

P

4

by the inverse of its molar mass.

6.2

g P

4

×

1

mol P

4

123.896

g P

4

=

0.05004 mol P

4

Moles of Phosphane

Multiply mol  

P

4

by the mole ratio,in the balanced equation, between  

P

4

and  

PH

3

that will cancel  

P

4

.

0.05004

mol P

4

×

4

mol PH

3

1

mol P

4

=

0.20016 mole PH

3

Mass of Phosphane

Multiply mol  

PH

3

by its molar mass.

0.20016

mol PH

3

×

33.998

g PH

3

1

mol PH

3

=

6.8 g PH

3

(rounded to two sig figs due to 6.2 g and 4.0 g)

Now you need to determine the mass of phosphane that  

4.0 g H

2

can produce. I'm going to put the steps together into one equation, but it will contain all of the steps required.

4.0

g H

2

×

1

mol H

2

2.016

g H

2

×

4

mol PH

3

6

mol H

2

×

33.998

g PH

3

1

mol PH

3

=

45 g PH

3

Phosphorus is the limiting reactant, which means the maximum amount of phosphane that can be produced is  

6.8 g

sorry its so long.

6.8 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.  

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

Step 1. Gather all the information in one place with molar masses above the formulas and masses below them.  

M_r:      123.90    2.016    34.00

               P₄   +     6H₂ ⟶ 4PH₃

Mass/g:  6.2         4.0

Step 2. Calculate the moles of each reactant  

Moles of P₄ = 6.2 × 1/123.9

Moles of P₄= 0.0500 mol P₄

Moles of H₂ = 4.2 × 1/2.016

Moles of H₂ = 2.08 mol H₂  

Step 3. Identify the limiting reactant  

Calculate the moles of PH₃ we can obtain from each reactant.  

From P₄ :

The molar ratio is 4 mol PH₃:1 mol P₄

Moles of PH₃ = 0.0500 × 4/1

Moles of PH₃= 0.200 mol PH₃

From H₂:

The molar ratio is 4 mol PH₃:6 mol H₂

Moles of PH₃= 2.08 × 4/6

Moles of PH₃ = 1.39 mol PH₃

The limiting reactant is P₄ because it gives the smaller amount of PH₃.

Step 4. Calculate the theoretical yield of PH₃ that you can obtain from P₄.

Theoretical yield of PH₃ = 0.200 × 34.00/1

Theoretical yield of PH₃ = 6.8 g PH₃

it is the most commonly accepted one

most historians agree that gabriel mouton, the vicar of st. paul’s church in lyons, france, is the “founding father” of the metric system. he proposed a decimal system of measurement in 1670. mouton based it on the length of one minute of arc of a great circle of the earth (now called a nautical mile, 1852 meters). he also proposed the swing-length of a pendulum with a frequency of one beat per second as the unit of length (about 25 cm). a pendulum beating with this length would have been fairly easy to produce, thus facilitating the widespread distribution of uniform standards. over the years, his work was revised, improved, and extended by a number of french scientists.

the political sponsor of weights and measures reform in the french revolutionary national assembly was the bishop of autun, better known as talleyrand. under his auspices, the french academy appointed several committees to carry out the work of developing a usable system of weights and measures for france. one of the committees recommended a decimalized measurement system based upon a length equal to one ten-millionth of the length of a quadrant of the earth’s meridian (i.e., one ten-millionth of the distance between the equator and the north pole).