# A Few Questions from a Worksheet on Stoichiometry

I'm just bringing myself back up to speed here, by over explaining what I'm doing to solve each question.

## #1

What mass of solid residue can be obtained from the thermal decomposition of 4.10g of anhydrous calcium nitrate?

Our equation is: $$Ca(NO_3)_2 \rightarrow CaO + 2NO_2 + O_2$$

Our solid residue is calcium oxide, since $$NO_2$$ and oxygen are gaseous. Calcium nitrate and calcium oxide have the same molar ratio. Hence, if we can deduce the number of moels of calcium nitrate, we can get the number of moles of calcium oxide.

The $$M_r$$ of calcium nitrate is $$40.1+2(14+(16\times3))$$ which gives us 164.1. The mass of calcium nitrate in this equation is 4.10g. Hence, with $$moles=\frac{mass}{M_r}$$, we get $$\frac{41}{1641}$$ moles.

With the molar ratio in mind, we know that $$(\frac{41}{1641})=\frac{mass}{40.1+16}$$. This gives us around about 1.399g.

Our options are (a) 0.70g, (b) 1.00g, (c) 1.40g, (d) 2.25g. C is our correct answer.

## #2

The combustion of fossil fuels is a major source of increasing atmospheric carbon dioxide, with a consequential rise in global warming. Another significant contribution to carbon dioxide levels comes from the thermal decomposition of limestone, in the manufacture of cement and of lime for agricultural purposes.

Cement works roat 1000 million tonnes of limestone per year and a further 200 million tonnes is roasted in kilns to make lime.

What is the total annual mass output of carbon dioxide (in million tonnes) from these two processes?

For simplicity, let's treat 1000+200=1200 million tonnes as grams. Our chemical equation is $$CaCO_3 \rightarrow CaO + CO_2$$.

The molar ratio between calcium carbonate and carbon dioxide is 1:1. Hence, let's calculate the moles of calcium carbonate. $$mol=\frac{1200}{40.1+12+3(16)}$$, or about 11.98 moles, whose non-rounded value we shall represent with the letter $$A$$, since this is non-rational and non-terminating decimal.

Remember the molar ratio. $$A=\frac{mass}{12+2(16)}$$. This gives us 527.4725... grams (tonnes).

Our options are (a) 440, (b) 527, (c) 660, (d) 880. B is our answer.

## #3

In an experiment, 0.125mol of chlorine gas, $$Cl_2$$, is reacted with an excess of cold aqueous sodium hydroxide. One of the products is a compound of sodium, oxygen, and chlorine.

Which mass of this product is formed?

A few things to note:

1. $$Cl_2+2NaOH\ (cold) \rightarrow NaCl+NaClO+H_2O$$
2. $$3Cl_2+6NaOH\ (hot) \rightarrow 5NaCl + NaClO_3 + H_2O$$

The first equation suits our purposes.

The molar ratio between chlorine and sodium chlorate is 1:1. The $$M_r$$ of sodium chlorate is 74.5. Multiplying this with the moles given (0.125) will give us our mass. In this case, 9.3125.

Our options are (a) 9.31g, (b) 13.3g, (c) 18.6g, (d) 26.6g. A is our answer.

## #4

The reaction between aluminium powder and anhydrous barium nitrate is used as the propellant in some fireworks. The reaction produces the metal oxides and nitrogen.

$$10Al+3Ba(NO_3)_2\rightarrow5Al_2O_3+3BaO+3N_2$$

Which mass of barium oxide is produced when 5.40g of aluminium power reacts with an excess of anhydrous barium nitrate?

Moles of aluminium: $$moles=\frac{5.40}{10(27)}$$ or 0.2.

The molar ratio of aluminium to barium oxide here is 10:3. Hence, to get the number of moles of barium oxide, we divide 0.2 with 10 and then multiply with 3. This gives us 0.06 moles. Now we can deduce the mass of barium oxide.

The $$M_r$$ of barium oxide here is $$(137.3+16)$$ or 153.3. This gives us about 9.198g.

Our options are (a) 1.62g, (b) 3.06g, (c) 9.18g, (d) 10.2g. Our answer is C.

## #5

Most modern cars are fitted with airbags. These work by decomposing sodium azide to liberate nitrogen gas, which inflates the bag.

$$2NaN_3 \rightarrow 3N_2 + 2Na$$

A typical driver' s airbag contains 50g of sodium azide.

Calculate the volume of nitrogen this will produce at room temperature.

Moles of sodium azide: $$\frac{50}{23+3(14)}$$ or $$\frac{10}{13}$$.

Molar ratio of sodium azide to nitrogen gas: 2:3.

$$M_r$$ of nitrogen gas: 28.

Hence, $$\frac{\frac{10}{13}}{2}\times3$$ or $$\frac{15}{13}$$ moles of nitrogen gas.

An error I made here was plugging these values into the mole/mass relationship. The question asks for the volume of nitrogen, not mass!

Thus, $$\frac{15}{13}\times24$$ gives us $$\frac{360}{13}$$, or around about 27.69 ($$moles=\frac{volume\ of\ gas\ in\ dm^3}{24}$$).

Our options are (a) 9.2$$dm^3$$, (b) 13.9$$dm^3$$, (c) 27.7$$dm^3$$, (d) 72.0$$dm^3$$. Our answer is C.

## #6

Which mass of gas would occupy a volume of 3$$dm^3$$ at 25 $$^{\circ}$$C and 1 atmosphere pressure? [1 mol of gas occupies 24$$dm^3$$ at 25 $$^{\circ}$$C and 1 atmosphere pressure.]

Our options are (a) 3.2g $$O_2$$ gas, (b) 5.6g $$N_2$$ gas, (c) 8.0g $$SO_2$$ gas, (d) 11.0g $$CO_2$$ gas.

We've got to work through most, if not all of these. Our procedure is to first convert grams to moles, then multiply that by 24. That will give us our volume.

$$O_2$$: $$\frac{3.2}{32} \rightarrow 0.1\times24 \rightarrow 2.4$$. Not oxygen gas.

$$N_2$$: $$\frac{5.6}{28} \rightarrow 0.2\times24 \rightarrow 4/8$$. Not nitrogen gas.

$$SO_2$$: $$\frac{8}{32.1+2(16)} \rightarrow \frac{80}{641}\times24 \rightarrow 2.9953...$$. Probably.

Let's do the last for confirmation.

$$CO_2$$: $$\frac{11}{12+2(16)} \rightarrow 0.25\times24 \rightarrow 3.5$$. Not this, then.

Our answer will be C, $$SO_2$$, then.

## #7

The reaction between aluminium powerder and anhydrous barium nitrate is used as the propellant in some fireworks. The metal oxides and nitrogen are the only products.

Which volume of nitrogen, measured under room conditions, is produced when 0.783g of anhydrous barium nitrate reacts with an excess of aluminium?

From the information in the question, we can deduce that our equation is something like $$Al + Ba(NO_3)_2 \rightarrow Al_2O_3+BaO+N_2$$ unbalanced. Balancing this equation requires more time than the effort it's worth.

The half-equation from here that concerns us is $$Ba(NO_3)_2 \rightarrow N_2$$. We can see that nitrogen is already balanced here. So this works. There is a 1:1 molar ratio.

Moles of barium nitrate: $$\frac{0.783}{137.3+2(14+3(16))} \rightarrow \frac{261}{87100}$$ moles.

Molar ratio: 1:1.

Hence, $$\frac{261}{87100}\times24000\rightarrow\frac{62640}{871}\approx71.917\ cm^3$$ ($$cm^3$$ because the answers are in that unit).

Our options are (a) 46.8 $$cm^3$$, (b) 72.0 $$cm^3$$, (c) 93.6 $$cm^3$$, (d) 144 $$cm^3$$. B is our answer.

## #8

The foul smell that skunks spray is due to a number of thiols, one of which is methanethiol, $$CH_3SH$$, which burns as follows:

$$CH_3SH+3O_2 \rightarrow CO_2+SO_2+2H_2O$$

A sample of 10 $$cm^3$$ of methanethiol was exploded with 60 $$cm^3$$ of oxygen.

What would be the final volume of the resultant mixture of gases when cooled to room temperature?

The ratio of moles is the same as the ratio of volume between reactants for gases only (R.T.P. does not matter here).

With this in mind, we can see that the molar ratio between $$CH_3SH$$ and $$3O_2$$ is 1:3. Using the $$10\ cm^3$$ provided, we know that $$30\ cm^3$$ of the oxygen will be used, leaving behind another $$30\ cm^3$$.

Since the molar ratio between $$CH_3SH$$, $$CO_2$$ and $$SO_2$$ is 1:1:1, we know that $$10\ cm^3$$ of carbon and sulfur dioxide is produced.

The resultant mixture of gases includes the volume of the products along with the volume of the unreacted oxygen. This sums to 10+10+30=$$50\ cm^3$$.

Our options are (a) 20 $$cm^3$$, (b) 30 $$cm^3$$, (c) 50 $$cm^3$$, (d) 70 $$cm^3$$. C is our answer here.

## #9

What volume of oxygen, measured under room conditions, can be obtained from the thermal decomposition of 8.2g of calcium nitrate ($$M_r$$=164)?

Our chemical equation is $$2Ca(NO_3)_2 \rightarrow 2CaO + 4NO_2 + O_2$$. The molar ratio of calcium nitrate to oxygen is 2:1, meaning we have to halve the moles of calcium nitrate.

There are $$\frac{8.2}{164} \rightarrow 0.05$$ moles of calcium nitrate. Half of this is 0.025 moles.

Our options are in $$cm^3$$, thus we shall multiply 0.025 with 24000. This gives us 600 $$cm^3$$.

The options are (a) 150 $$cm^3$$, (b) 300 $$cm^3$$, (c) 600 $$cm^3$$, (d) 1200 $$cm^3$$. C is our answer.

## The End

I think I've gotten back up to speed. Still could use more work, but good enough thus far.

This article was written on 16/09/2023. If you have any thoughts, feel free to send me an email with them. Have a nice day!