Question

1. how many grams of o₂ will be formed from 3.76 grams of kclo₃?

\\(\frac{3.76\text{g }kclo_3}{}\vert\frac{1\text{ mol }kclo_3}{122.55\text{ g }kclo_3}\vert\frac{\\_\\_\text{ mol }o_2}{\\_\\_\text{ mol }kclo_3}\vert\frac{\\_\\_\text{ g }o_2}{\\_\\_\text{ mol }o_2}\vert=\\_\\_\\_\text{ g }o_2\\)

1. how many grams of kclo₃ are needed to make 30.0 grams of kcl?

\\(\frac{30.0\text{ g }kcl}{}\vert\frac{\\_\\_\\_\text{ mol }kcl}{\\_\\_\\_\text{ g }kcl}\vert\frac{\\_\\_\\_\text{ mol }kclo_3}{\\_\\_\\_\text{ mol }kcl}\vert\frac{\\_\\_\\_\text{ g }kclo_3}{\\_\\_\\_\text{ mol }kclo_3}\vert=\\_\\_\\_\\_\text{ g }kclo_3\\)

1. how many grams of kcl will be formed from 2.73 g of kclo₃?

\\(\frac{2.73\text{ g }kclo_3}{}\vert\frac{}{}\vert\frac{}{}\vert\frac{}{}\vert=\\_\\_\\_\\_\text{ g }kcl\\)
\\(4fe + 3o_2 \
ightarrow 2fe_2o_3\\)

1. how many grams of fe₂o₃ are produced when 42.7 grams of fe is reacted?

\\(\frac{42.7\text{ g }fe}{}\vert\frac{\\_\\_\\_\text{ mol }fe}{\\_\\_\\_\text{ g }fe}\vert\frac{\\_\\_\\_\text{ mol }fe_2o_3}{\\_\\_\\_\text{ mol }fe}\vert\frac{\\_\\_\\_\text{ g }fe_2o_3}{\\_\\_\\_\text{ mol }fe_2o_3}\vert=\\_\\_\\_\\_\text{ g }fe_2o_3\\)

1. how many grams of fe₂o₃ are produced when 17.0 grams of o₂ is reacted?

\\(\frac{17.0\text{ g }o_2}{}\vert\frac{}{}\vert\frac{}{}\vert\frac{}{}\vert=\\_\\_\\_\\_\text{ g }fe_2o_3\\)

1. how many grams of o₂ are needed to react with 125 grams of fe?

\\(\frac{125\text{g }fe}{}\vert\frac{1\text{ mol }fe}{56\text{ g }fe}\vert\frac{3\text{ mol }o_2}{4\text{ mole }fe}\vert\frac{32\text{ g }o_2}{1\text{ mol }o_2}\vert=53.57\text{ g }o_2\\)
some cars can use butane (c₄h₁₀) as fuel:
\\(2c_4h_{10} + 13o_2 \
ightarrow 8co_2 + 10h_2o\\)

1. how many grams of co₂ are produced from the combustion of 100. grams of butane?

\\(\frac{100.\text{ g }c_4h_{10}}{}\vert\frac{}{}\vert\frac{}{}\vert\frac{}{}\vert=\\_\\_\\_\\_\text{ g }co_2\\)

1. how many grams of o₂ are needed to react with of 100. grams of butane?

\\(\frac{100.\text{ g }c_4h_{10}}{}\vert\frac{}{}\vert\frac{}{}\vert\frac{}{}\vert=\\_\\_\\_\\_\text{ g }o_2\\)
18 how many grams of h₂o are produced when 5.38g of o₂ is reacted?

Explanation:

First, we need the balanced chemical equation for the decomposition of $\text{KClO}_3$:

$$2\text{KClO}_3 ightarrow 2\text{KCl} + 3\text{O}_2$$

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Problem 10

Step1: Convert mass to moles of $\text{KClO}_3$

$$3.76\ \text{g KClO}_3 \times \frac{1\ \text{mol KClO}_3}{122.55\ \text{g KClO}_3}$$

Step2: Mole ratio to $\text{O}_2$

$$\times \frac{3\ \text{mol O}_2}{2\ \text{mol KClO}_3}$$

Step3: Convert moles to mass of $\text{O}_2$

$$\times \frac{32.00\ \text{g O}_2}{1\ \text{mol O}_2}$$

Step4: Calculate final value

$$3.76 \times \frac{1}{122.55} \times \frac{3}{2} \times 32.00 = 1.48$$

Step1: Convert mass to moles of $\text{KCl}$

$$30.0\ \text{g KCl} \times \frac{1\ \text{mol KCl}}{74.55\ \text{g KCl}}$$

Step2: Mole ratio to $\text{KClO}_3$

$$\times \frac{2\ \text{mol KClO}_3}{2\ \text{mol KCl}}$$

Step3: Convert moles to mass of $\text{KClO}_3$

$$\times \frac{122.55\ \text{g KClO}_3}{1\ \text{mol KClO}_3}$$

Step4: Calculate final value

$$30.0 \times \frac{1}{74.55} \times \frac{2}{2} \times 122.55 = 49.4$$

Step1: Convert mass to moles of $\text{KClO}_3$

$$2.73\ \text{g KClO}_3 \times \frac{1\ \text{mol KClO}_3}{122.55\ \text{g KClO}_3}$$

Step2: Mole ratio to $\text{KCl}$

$$\times \frac{2\ \text{mol KCl}}{2\ \text{mol KClO}_3}$$

Step3: Convert moles to mass of $\text{KCl}$

$$\times \frac{74.55\ \text{g KCl}}{1\ \text{mol KCl}}$$

Step4: Calculate final value

$$2.73 \times \frac{1}{122.55} \times \frac{2}{2} \times 74.55 = 1.66$$

Answer:

$1.48$ g $\text{O}_2$

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Problem 11