I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
Report (any type)/Brief report
22/09/2021
This is a long question. It has a long answer.
I must point out that all the substances in the equation are soluble in water, so there is actually No Reaction.
If there were a reaction, this is how you would attack the problem.
BALANCE THE EQUATION:
Count the ##”NH”_4##, ##”NO”_3##, and ##”PO”_4## as single units.
Start with the most complicated formula — ##”(NH”_4″)”_3″PO”_4##. Put a 1 in front of it.
##”NH”_4″NO”_3 + “Na”_3″PO”_4 → color(red)(1)(“NH”_4”)”_3″PO”_4 + “NaNO”_3##
Then balance ##”NH”_4##. Put a 3 in front of ##”NH”_4″NO”_3##.
## color(teal)(3)”NH”_4″NO”_3 + “Na”_3″PO”_4 → color(red)(1)(“NH”_4”)”_3″PO”_4 + “NaNO”_3##
Balance ##”PO”_4##. Put a 1 in front of ##”Na”_3″PO”_4##.
## color(teal)(3)”NH”_4″NO”_3 + color(blue)(1)”Na”_3″PO”_4 → color(red)(1)(“NH”_4)_3″PO”_4 + “NaNO”_3##
Balance ##”Na”##. Put a 3 in front of ##”NaNO”_3##.
## color(teal)(3)”NH”_4″NO”_3 + color(blue)(1)”Na”_3″PO”_4 → color(red)(1)(“NH”_4)_3″PO”_4 + color(orange)(3)”NaNO”_3##
The balanced equation is
##color(red)(“3NH”_4″NO”_3 + “Na”_3″PO”_4 → “(NH”_4”)”_3″PO”_4 + “3NaN”O_3)##
IDENTIFY THE LIMITING REACTANT
The molar masses are
##”NH”_4″NO”_3= “80.04 g/mol”##; ##”Na”_3″PO”_4 = “163.94 g/mol”##
##”(NH”_4″)”_3″PO”_4 = “149.09 g/mol”##; ##”NaNO”_3 = “84.99 g/mo”l##
Mass of ##”(NH”_4″)”_3″PO”_4## from ##”NH”_4″NO”_3##:
##30.0 cancel(“g NH₄NO₃”) × (1 cancel(“mol NH₄NO₃”))/(80.04 cancel(“g NH₄NO₃”)) × (1 cancel(“mol (NH₄)₃PO₄”))/(3 cancel(“mol NH₄NO₃”)) ×##
##(“149.09 g (NH”_4”)”_3″PO”_4)/(1 cancel(“mol (NH₄)₃PO₄”)) = “18.6 g (NH”_4”)”_3″PO”_4##
Mass of ##(“NH”_4”)”_3″PO”_4## from ##”Na”_3″PO”_4##:
##50.0 cancel(“g Na₃PO₄”)× (1 cancel(“mol Na₃PO₄”))/(163.94 cancel(“g Na₃PO₄”)) × (1 cancel(“mol (NH₄)₃PO₄”))/(1 cancel(“mol Na₃PO₄”)) ×##
##(“149.09 g (NH”_4”)”_3″PO”_4)/(1 cancel(“mol (NH₄)₃PO₄”)) = “45.5 g (NH”_4”)”_3″PO”_4##
##”NH”_4″NO”_3## is the limiting reactant, because it gives the smaller amount of product.
MASS OF EACH PRODUCT
Mass of ##(“NH”_4”)”_3″PO”_4 = “18.6 g”##
Mass of ##”NaNO”_3##:
##30.0 cancel(“g NH₄NO₃”) × (1 cancel(“mol NH₄NO₃”))/(80.04 cancel(“g NH₄NO₃”)) × (3 cancel(“mol NaNO₃”))/(3 cancel(“mol NH₄NO₃”)) ×##
##(“84.99 g NaNO”_3)/(1 cancel(“mol NaNO₃”)) = “31.9 g NaNO”_3##
MASS OF LIMITING REACTANT REMAINING
The mass of ##”NH”_4″NO”_3## remaining is zero, since the reaction uses up all the limiting reactant.
MASS OF EXCESS REACTANT REMAINING
Mass of ##”Na”_3″PO”_4## reacted:
##30 cancel(“g NH₄NO₃”) × (1 cancel(“mol NH₄NO₃”))/(80.04 cancel(“g NH₄NO₃”)) × (1 cancel(“mol Na₃PO₄”))/(3 cancel(“mol NH₄NO₃”)) ×##
##(“163.94 g Na”_3″PO”_4)/(1 cancel(“mol Na₃PO₄”)) = “20.5 g (NH”_4”)”_3″PO”_4##
The mass of ##”Na”_3″PO”_4 ” remaining = 50.0 g – 20.5 g = 29.5 g”##
I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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