decomposition of potassium chlorate (kclo3) produces potassium chloride (kcl) and pure oxygen (o2). the…

decomposition of potassium chlorate (kclo3) produces potassium chloride (kcl) and pure oxygen (o2). the balanced equation for the reaction is as follows. 2kclo3(s)→2kcl(s)+3o2(g) what volume of oxygen gas is released at stp if 10.0 g of potassium chlorate is decomposed? (the molar mass of kclo3 is 122.55 g/mol.) 0.914 l 1.83 l 2.74 l 3.66 l

decomposition of potassium chlorate (kclo3) produces potassium chloride (kcl) and pure oxygen (o2). the balanced equation for the reaction is as follows. 2kclo3(s)→2kcl(s)+3o2(g) what volume of oxygen gas is released at stp if 10.0 g of potassium chlorate is decomposed? (the molar mass of kclo3 is 122.55 g/mol.) 0.914 l 1.83 l 2.74 l 3.66 l

Answer

Explanation:

Step1: Calculate moles of KClO₃

$n_{KClO_3}=\frac{m}{M}=\frac{10.0\ g}{122.55\ g/mol}\approx0.0816\ mol$

Step2: Determine moles of O₂ from stoichiometry

From the balanced equation $2KClO_3(s)\longrightarrow2KCl(s) + 3O_2(g)$, the mole - ratio of $KClO_3$ to $O_2$ is 2:3. So $n_{O_2}=\frac{3}{2}n_{KClO_3}=\frac{3}{2}\times0.0816\ mol = 0.1224\ mol$

Step3: Calculate volume of O₂ at STP

At STP, 1 mole of any gas occupies 22.4 L. So $V_{O_2}=n_{O_2}\times22.4\ L/mol=0.1224\ mol\times22.4\ L/mol\approx2.74\ L$

Answer:

2.74 L