Start with a balanced equation.
#"P"_4 + "6H"_2"##rarr##"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:
#color(blue)("given mass P"_4"##rarr##color(blue)("mole P"_4"##rarr##color(red)("mol PH"_3"##rarr##color(purple)("mass PH"_3"#
and
#color(blue)("given mass H"_2"##rarr##color(blue)("mole H"_2"##rarr##color(red)("mol PH"_3"##rarr##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.
#"P"_4:##(4xx30.974"g/mol P")="123.896 g/mol P"_4"#
#"H"_2:##(2xx1.008"g/mol H")="2.016 g/mol H"_2"#
#"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, #"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.
#color(blue)("Moles of Phosphorus"#
Multiply the given mass of #"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 #"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"#.
