#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