Question

10 numeric 10 points 1.50 g of nitrogen monoxide react with 2.00 g of oxygen gas to produce nitrogen trioxide according to equation 1. equation 1. no (g) + o₂ (g) → no₃ (g) after all of the no has reacted, the excess o₂ is then reacted with 1.00 g of br₂ to produce bromine monoxide according to equation 2. equation 2. o₂ (g) + br₂ (l) → 2 bro (l) calculate how many grams of bro will be produced. answer

Explanation:

Step1: Balance the first - reaction equation

The balanced equation for the reaction of nitrogen monoxide and oxygen is \(2NO(g)+O_2(g)
ightarrow 2NO_2(g)\).
The molar mass of \(NO\) is \(M_{NO}=14 + 16=30\ g/mol\), and the molar mass of \(O_2\) is \(M_{O_2}=2\times16 = 32\ g/mol\).
The number of moles of \(NO\), \(n_{NO}=\frac{m_{NO}}{M_{NO}}=\frac{1.50\ g}{30\ g/mol}=0.05\ mol\).
From the balanced equation, the moles of \(O_2\) required to react with \(NO\) is \(n_{O_2 - required - for - NO}=\frac{1}{2}n_{NO}=\frac{1}{2}\times0.05\ mol = 0.025\ mol\).
The mass of \(O_2\) required to react with \(NO\) is \(m_{O_2 - required - for - NO}=n_{O_2 - required - for - NO}\times M_{O_2}=0.025\ mol\times32\ g/mol = 0.8\ g\).
The initial mass of \(O_2\) is \(m_{O_2 - initial}=2.00\ g\), so the excess mass of \(O_2\) is \(m_{O_2 - excess}=2.00\ g - 0.8\ g=1.2\ g\).
The number of moles of \(O_2\) in the excess amount is \(n_{O_2 - excess}=\frac{m_{O_2 - excess}}{M_{O_2}}=\frac{1.2\ g}{32\ g/mol}=0.0375\ mol\).

Step2: Balance the second - reaction equation

The balanced equation for the reaction of \(O_2\) and \(Br_2\) is \(O_2(g)+2Br_2(l)
ightarrow 4BrO(l)\).
The molar mass of \(Br_2\) is \(M_{Br_2}=2\times79.9 = 159.8\ g/mol\), and the number of moles of \(Br_2\) is \(n_{Br_2}=\frac{m_{Br_2}}{M_{Br_2}}=\frac{1.00\ g}{159.8\ g/mol}\approx0.00626\ mol\).
From the balanced equation, the moles of \(O_2\) required to react with \(Br_2\) is \(n_{O_2 - required - for - Br_2}=\frac{1}{2}n_{Br_2}=\frac{1}{2}\times0.00626\ mol = 0.00313\ mol\). Since \(n_{O_2 - excess}=0.0375\ mol\gt n_{O_2 - required - for - Br_2}\), \(Br_2\) is the limiting reactant.

Step3: Calculate the moles and mass of \(BrO\)

From the balanced equation \(O_2(g)+2Br_2(l)
ightarrow 4BrO(l)\), the mole - ratio of \(Br_2\) to \(BrO\) is \(1:2\).
The number of moles of \(BrO\) produced is \(n_{BrO}=2n_{Br_2}=2\times0.00626\ mol = 0.01252\ mol\).
The molar mass of \(BrO\) is \(M_{BrO}=79.9+16 = 95.9\ g/mol\).
The mass of \(BrO\) produced is \(m_{BrO}=n_{BrO}\times M_{BrO}=0.01252\ mol\times95.9\ g/mol\approx1.19\ g\).

Answer:

\(1.19\ g\)