2 kclo₃ → 2 kcl + 3 o₂\n4) given 33.79 grams of o₂, how many moles of kclo₃ are produced?\n5) given 2.8…

4)

Explanation:

Step1: Calculate moles of O₂

The molar - mass of O₂ is (M_{O_2}=2\times16\ g/mol = 32\ g/mol). The number of moles of O₂, (n_{O_2}=\frac{m_{O_2}}{M_{O_2}}), where (m_{O_2} = 33.79\ g). So (n_{O_2}=\frac{33.79\ g}{32\ g/mol}=1.056\ mol).

Step2: Use mole - ratio from the balanced equation

From the balanced equation (2KClO_3\rightarrow2KCl + 3O_2), the mole - ratio of (KClO_3) to (O_2) is (\frac{n_{KClO_3}}{n_{O_2}}=\frac{2}{3}). Then (n_{KClO_3}=\frac{2}{3}n_{O_2}). Substituting (n_{O_2}=1.056\ mol), we get (n_{KClO_3}=\frac{2}{3}\times1.056\ mol = 0.704\ mol).

Answer:

(0.704\ mol)

5)

Explanation:

Step1: Use mole - ratio from the balanced equation

From the balanced equation (2KClO_3\rightarrow2KCl + 3O_2), the mole - ratio of (KCl) to (O_2) is (\frac{n_{KCl}}{n_{O_2}}=\frac{2}{3}). Given (n_{O_2}=2.8\ mol), then (n_{KCl}=\frac{2}{3}n_{O_2}=\frac{2}{3}\times2.8\ mol=\frac{5.6}{3}\ mol).

Step2: Calculate mass of KCl

The molar - mass of KCl is (M_{KCl}=39.1 + 35.5=74.6\ g/mol). The mass of KCl, (m_{KCl}=n_{KCl}\times M_{KCl}). Substituting (n_{KCl}=\frac{5.6}{3}\ mol) and (M_{KCl}=74.6\ g/mol), we get (m_{KCl}=\frac{5.6}{3}\ mol\times74.6\ g/mol\approx138.45\ g).

Answer:

(138.45\ g)

6)

Explanation:

Step1: Calculate moles of KClO₃

The molar - mass of (KClO_3) is (M_{KClO_3}=39.1+35.5 + 3\times16=122.6\ g/mol). The number of moles of (KClO_3), (n_{KClO_3}=\frac{m_{KClO_3}}{M_{KClO_3}}), where (m_{KClO_3}=6.25\ g). So (n_{KClO_3}=\frac{6.25\ g}{122.6\ g/mol}\approx0.051\ mol).

Step2: Use mole - ratio from the balanced equation

From the balanced equation (2KClO_3\rightarrow2KCl + 3O_2), the mole - ratio of (KCl) to (KClO_3) is (\frac{n_{KCl}}{n_{KClO_3}}=\frac{2}{2}=1). So (n_{KCl}=n_{KClO_3}=0.051\ mol).

Step3: Calculate mass of KCl

The molar - mass of KCl is (M_{KCl}=74.6\ g/mol). The mass of KCl, (m_{KCl}=n_{KCl}\times M_{KCl}). Substituting (n_{KCl}=0.051\ mol) and (M_{KCl}=74.6\ g/mol), we get (m_{KCl}=0.051\ mol\times74.6\ g/mol\approx3.80\ g).

Answer:

(3.80\ g)