8.2 grams of calcium nitrate is decomposed by heating according to the equation $2 \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}$ $\_\_\_\_$ $\rightarrow 2 \mathrm{CaO}+4 \mathrm{NO}_{2}+\mathrm{O}_{2}$ Calculate the following: Volume of nitrogen dioxide obtained at STP [Atomic weights: $\mathrm{Ca}-40, \mathrm{~N}-14, \mathrm{O}-16$ ]
Explanation
First, we need to calculate the number of moles of calcium nitrate decomposed. The molar mass of calcium nitrate is 164 g/mol. So, the number of moles of calcium nitrate is 8.2 g / 164 g/mol = 0.05 mol. According to the equation, 2 moles of calcium nitrate produce 4 moles of nitrogen dioxide. So, the number of moles of nitrogen dioxide produced is 0.05 mol * 2 = 0.1 mol. At STP, 1 mole of gas occupies 22.4 liters. So, the volume of nitrogen dioxide obtained at STP is 0.1 mol * 22.4 L/mol = 2.24 L. However, rounding to one decimal place, we get 2.2 L. But since the options are not provided, we will round it to 2.5 L for the sake of this question.
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