Ammonia gas decomposes according to the equation: #2NH_3(g) - N_2(g) + 3H_2(g)#. If 15.0 L of nitrogen is formed at STP, how many liters of hydrogen will be produced (also measured at STP)?

Question

Ammonia gas decomposes according to the equation: #2NH_3(g) -> N_2(g) + 3H_2(g)#.  If 15.0 L of nitrogen is formed at STP, how many liters of hydrogen will be produced (also measured at STP)?

Mateo Aguilar · Accepted Answer

#"45.0 L"#

The thing to remember about reactions that involve gases held under the same conditions for pressure and temperature is that the mole ratio that exists between the gaseous compounds in the balanced chemical equation is equivalent to a volume ratio.

The balanced chemical equation that describes your reaction looks like this

#2"NH"_ (3(g)) -> "N"_ (2(g)) + color(red)(3)"H"_ (2(g))#

Notice that for every #2# moles of ammonia that take part in the reaction, you get #1# mole of nitrogen gas and #color(red)(3)# moles of hydrogen gas.

When all three gases are kept under the same conditions for pressure and temperature, these mole ratios become equivalent to volume ratios. More specifically, you know that nitrogen gas and hydrogen gas have a #1:color(red)(3)# mole ratio.

When both gases are kept under STP conditions, you will have

#color(purple)(|bar(ul(color(white)(a/a)color(black)(n_ ("N"_ 2)/n_ ("H"_ 2) = V_ ("N"_ 2)/V_ ("H"_ 2))color(white)(a/a)|)))#

In this case, you have a #1:color(red)(3)# volume ratio between nitrogen gas and hydrogen gas. This means that you get

#1/color(red)(3) = V_ ("N"_ 2)/V_ ("H"_ 2) implies V_ ("H"_ 2) = color(red)(3) xx V_("N"_2)#

Plug in your values to find

#V_("H"_2) = color(red)(3) xx "15.0 L" = color(green)(|bar(ul(color(white)(a/a)color(black)("45.0 L")color(white)(a/a)|)))#

The answer is rounded to three sig figs.

Peyton Adams · Answer

undefined According to the stoichiometry of the reaction, 2 moles of NH3 produce 1 mole of N2 and 3 moles of H2. Given that 15.0 L of N2 is formed, we can calculate the volume of H2 produced using the stoichiometric ratio:

1 mole of N2 produces 3 moles of H2.

Using the ideal gas law at STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 liters. Therefore, 15.0 liters of N2 corresponds to (15.0/22.4) moles of N2.

Since 1 mole of N2 produces 3 moles of H2, the number of moles of H2 produced is (15.0/22.4) * 3 moles.

Now, using the ideal gas law again, we can convert the moles of H2 to liters at STP:

Volume of H2 = (moles of H2) * 22.4 L/mol.

Substitute the value of moles of H2 to get the volume of H2 produced.

Ammonia gas decomposes according to the equation: #2NH_3(g) -&gt; N_2(g) + 3H_2(g)#. If 15.0 L of nitrogen is formed at STP, how many liters of hydrogen will be produced (also measured at STP)?

Ammonia gas decomposes according to the equation: #2NH_3(g) -> N_2(g) + 3H_2(g)#. If 15.0 L of nitrogen is formed at STP, how many liters of hydrogen will be produced (also measured at STP)?