#N_2# + 3#H_2# ---> 2N#H_3# What volume of hydrogen is necessary to react with five liters of nitrogen to produce ammonia? (Assume constant temperature and pressure.)
#N_2# + 3#H_2# ---> 2N#H_3# , as per this equation, 1 volume of Nitrogen , needs 3 volumes of Hydrogen. The reaction produces two volumes of ammonia.
Let us write this as equation;
#"1 volume of Nitrogen"/"3 volume of Hydrogen"# (a)
If we start with 5 L of Nitrogen , X volumes of Hydrogen will be needed.
Let us write this as equation;
#"5 L of Nitrogen"/"X L of Hydrogen"# (b)
equate (a) and (b)
#1/3# = #5/X#
cross multiply #1 * X# = #3 * 5# , X=15 L
2CO + #O_2# ---> 2C#O_2# How many liters of carbon dioxide are produced if 75 liters of carbon monoxide are burned in oxygen? How many liters of oxygen are necessary?
As per this equation, 2 volumes of Carbon Monoxide, needs 1 volume of Oxygen. The reaction produces two volumes of Carbon dioxide.
Let us write this as equation;
#"2 volume of Carbon Monoxide"/"2 volumes of Carbon Dioxide"# (a)
If we start with 75 L of Carbon monoxide , X volumes of carbon dioxide will be produced.
Let us write this as equation;
#"75 L of CO"/("X L of "CO_2)# (b)
equate (a) and (b)
#2/2# = #75/X#
cross multiply #2*X# = #2*75# , X= 75 L
#"2 volume of Carbon Monoxide"/"1 volumes of Oxygen"# (a)
If we start with 75 L of Carbon Monoxide , X volumes of Oxygen will be needed.
#"75 L of CO"/("X L of "O_2)# (b)
equate (a) and (b)
#2/1#= #75/X#
#2*X# = #75*1#
2X =75
X= 37.5 L
