Start with a balanced equation.
"P"_4 + "6H"_2"P4+6H2rarr→"4PH"_3"4PH3
This is a limiting reactant problem. The reactant that produces the least "PH"_3"PH3 determines the maximum number of grams of "PH"_3"PH3
The process will go as follows:
color(blue)("given mass P"_4"given mass P4rarr→color(blue)("mole P"_4"mole P4rarr→color(red)("mol PH"_3"mol PH3rarr→color(purple)("mass PH"_3"mass PH3
and
color(blue)("given mass H"_2"given mass H2rarr→color(blue)("mole H"_2"mole H2rarr→color(red)("mol PH"_3"mol PH3rarr→color(purple)("mass PH"_3"mass PH3
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"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:P4:(4xx30.974"g/mol P")="123.896 g/mol P"_4"(4×30.974g/mol P)=123.896 g/mol P4
"H"_2:H2:(2xx1.008"g/mol H")="2.016 g/mol H"_2"(2×1.008g/mol H)=2.016 g/mol H2
"PH"_3:PH3:(1xx30.974"g/mol P")+(3xx1.008"g/mol H")="33.998 g/mol PH"_3"(1×30.974g/mol P)+(3×1.008g/mol H)=33.998 g/mol PH3
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Now you need to determine the mass of phosphane, "PH"_3"PH3, produced individually by "6.2 g P"_46.2 g P4 and by "4.0 g H"_2"4.0 g H2. I'm going to start with "P"_4"P4 since it is first in the equation.
color(blue)("Moles of Phosphorus"Moles of Phosphorus
Multiply the given mass of "P"_4"P4 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 "P"_4 by the mole ratio,in the balanced equation, between "P"_4" and "PH"_3 that will cancel "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 "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 "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 "6.8 g".