Question

How many liters of NH3, at STP, will react with 19.5 g O2 to form NO and H2O?

Answer 1

Answer:

10.9 L

Explanation:

Let's begin by introducing the strategy to solve this problem:

• Write a chemical reaction and make sure it's balanced;
• Identify STP (standard temperature and pressure, that is, T = 273.15 K and p = 1.00 atm);
• Find moles of oxygen;
• From the stoichiometry of the equation, identify the number of moles of ammonia;
• Convert moles of ammonia into volume using the ideal gas law pV = nRT.

(a) The first step is already fulfilled: the reaction is balanced and it shows that 4 moles of ammonia react with 5 moles of oxygen.

(b) Given the following variables:

$p=1.00 atm\\T=273.15 K\\m_O_2=19.5 g\\R=0.08206 \frac{L\cdot atm}{mol\cdot K}$

(c) Find moles of oxygen dividing mass of oxygen by its molar mass:

$n_O_2=\frac{m_O_2}{M_O_2}$

Here molar mass of oxygen is:

$M_O_2=32.00 g/mol$

(d) From stoichiometry, dividing moles of ammonia by its stoichiometric coefficient would be equal to the ratio of moles of oxygen divided by its stoichiometric coefficient:

$\frac{n_{NH_3}}{4} =\frac{n_O_2}{5}$

Rearrange the equation to obtain moles of ammonia:

$n_{NH_3}=\frac{4}{5} n_O_2$

(e) Solve pV = nRT for volume of ammonia:

$pV_{NH_3}=n_{NH_3}RT\\\therefore V_{NH_3}=\frac{n_{NH_3}RT}{p}$

$V_{NH_3}=\frac{\frac{4}{5}n_O_2RT }{p} =\frac{\frac{4m_O_2}{5M_O_2}RT }{p}=\frac{4m_O_2RT}{5pM_O_2}$

Substitute the given data to obtain the final answer!

${V}=\frac{4m_O_2RT}{5pM_O_2} =\frac{4\cdot19.5 g\cdot0.08206 \frac{L\cdot atm}{mol\cdot K}\cdot273.15 K }{5\cdot 1.00 atm\cdot32.00 g/mol}= 10.9 L$