Questions: For the unbalanced reaction below, calculate how many grams of potassium chloride form when 7.09 g of potassium chlorate completely reacts. Be sure to balance the reaction, use significant figures and units in the answer. KClO3(s) --> KCl(s) + O2(g)

Solution Steps

First, we need to balance the given chemical equation: \[ \mathrm{KClO}_{3}(\mathrm{~s}) \rightarrow \mathrm{KCl}(\mathrm{~s}) + \mathrm{O}_{2}(\mathrm{~g}) \]

Balancing the equation, we get: \[ 2 \mathrm{KClO}_{3}(\mathrm{~s}) \rightarrow 2 \mathrm{KCl}(\mathrm{~s}) + 3 \mathrm{O}_{2}(\mathrm{~g}) \]

Next, we calculate the molar masses of potassium chlorate (KClO\(_3\)) and potassium chloride (KCl).

• Molar mass of KClO\(_3\): \[ \text{K} = 39.10 \, \text{g/mol}, \quad \text{Cl} = 35.45 \, \text{g/mol}, \quad \text{O}_3 = 3 \times 16.00 \, \text{g/mol} \] \[ \text{Molar mass of KClO}_3 = 39.10 + 35.45 + 48.00 = 122.55 \, \text{g/mol} \]

• Molar mass of KCl: \[ \text{K} = 39.10 \, \text{g/mol}, \quad \text{Cl} = 35.45 \, \text{g/mol} \] \[ \text{Molar mass of KCl} = 39.10 + 35.45 = 74.55 \, \text{g/mol} \]

Convert the given mass of KClO\(_3\) to moles: \[ \text{Moles of KClO}_3 = \frac{7.09 \, \text{g}}{122.55 \, \text{g/mol}} = 0.05787 \, \text{mol} \]

Using the balanced equation, we find the moles of KCl produced: \[ 2 \, \text{mol KClO}_3 \rightarrow 2 \, \text{mol KCl} \] \[ \text{Moles of KCl} = 0.05787 \, \text{mol KClO}_3 \times \frac{2 \, \text{mol KCl}}{2 \, \text{mol KClO}_3} = 0.05787 \, \text{mol KCl} \]

Convert the moles of KCl to grams: \[ \text{Mass of KCl} = 0.05787 \, \text{mol} \times 74.55 \, \text{g/mol} = 4.315 \, \text{g} \]

Final Answer