# What is the maximum number of grams of PH_3 that can be formed when 6.2 g of phosphorus reacts with 4.0 g of hydrogen to form PH_3?

May 23, 2017

#### Answer:

The maximum mass of phosphane $\left({\text{PH}}_{3}\right)$ that can be produced under the conditions stated in the question is $\text{6.8 g}$.

#### Explanation:

Start with a balanced equation.

$\text{P"_4 + "6H"_2}$$\rightarrow$$\text{4PH"_3}$

This is a limiting reactant problem. The reactant that produces the least $\text{PH"_3}$ determines the maximum number of grams of $\text{PH"_3}$

The process will go as follows:

color(blue)("given mass P"_4"$\rightarrow$color(blue)("mole P"_4"$\rightarrow$color(red)("mol PH"_3"$\rightarrow$color(purple)("mass PH"_3"

and

color(blue)("given mass H"_2"$\rightarrow$color(blue)("mole H"_2"$\rightarrow$color(red)("mol PH"_3"$\rightarrow$color(purple)("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.

color(green)("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.

${\text{P}}_{4} :$(4xx30.974"g/mol P")="123.896 g/mol P"_4"

${\text{H}}_{2} :$(2xx1.008"g/mol H")="2.016 g/mol H"_2"

${\text{PH}}_{3} :$(1xx30.974"g/mol P")+(3xx1.008"g/mol H")="33.998 g/mol PH"_3"

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Now you need to determine the mass of phosphane, $\text{PH"_3}$, produced individually by ${\text{6.2 g P}}_{4}$ and by $\text{4.0 g H"_2}$. I'm going to start with $\text{P"_4}$ since it is first in the equation.

color(blue)("Moles of Phosphorus"
Multiply the given mass of $\text{P"_4}$ by the inverse of its molar mass.

6.2color(red)cancel(color(black)("g P"_4))xx(1"mol P"_4)/(123.896color(red)cancel(color(black)("g P"_4)))="0.05004 mol P"_4

color(red)("Moles of Phosphane"
Multiply mol ${\text{P}}_{4}$ by the mole ratio,in the balanced equation, between $\text{P"_4}$ and ${\text{PH}}_{3}$ that will cancel $\text{P"_4}$.

0.05004color(red)cancel(color(black)("mol P"_4))xx(4"mol PH"_3)/(1color(red)cancel(color(black)("mol P"_4)))="0.20016 mole PH"_3"

color(purple)("Mass of Phosphane"
Multiply mol ${\text{PH}}_{3}$ by its molar mass.

0.20016color(red)cancel(color(black)("mol PH"_3))xx(33.998"g PH"_3)/(1color(red)cancel(color(black)("mol PH"_3)))="6.8 g PH"_3" (rounded to two sig figs due to 6.2 g and 4.0 g)

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Now you need to determine the mass of phosphane that ${\text{4.0 g H}}_{2}$ can produce. I'm going to put the steps together into one equation, but it will contain all of the steps required.

color(black)cancel(color(blue)(4.0"g H"_2))xxcolor(blue)((1color(black)cancel(color(blue)("mol H"_2)))/(2.016color(black)cancel(color(blue)("g H"_2)))xxcolor(red)((4color(black)cancel(color(red)("mol PH"_3)))/(6color(black)cancel(color(red)("mol H"_2)))xxcolor(purple)((33.998"g PH"_3)/(1color(black)cancel(color(purple)("mol PH"_3)))=color(purple)("45 g PH"_3

Phosphorus is the limiting reactant, which means the maximum amount of phosphane that can be produced is $\text{6.8 g}$.