10 numeric 10 points 1.50 g of nitrogen monoxide react with 2.00 g of oxygen gas to produce nitrogen…

Explanation:

Step1: Balance the first - reaction equation

The balanced equation for the reaction of nitrogen monoxide and oxygen is (2NO(g)+O_2(g)\rightarrow 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)\rightarrow 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)\rightarrow 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)