Hi, Zoe,

Warning - I got most of the way through, but have to leave and did not complete the last problem. This should help, however.

Bob

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1) Sulfur burns in excess air to form sulfur dioxide according to the equation: S(s)+O2(g)-> SO2(g)

What volume of sulfur dioxide is produced (at room temperature and pressure) from 24g of sulfur.

The equation is balanced, and it tells us that we will get 1 mole of SO2 for every mole of S reacted. Calculate the number of moles of sulfur in 24g. The molar mass of S is 32.1. Therefore, 24g/(32.1g/mole) = 0.748 moles (I will ignore sig figs until the end). That means we get 0.748 moles of SO2 produced, assuming the O2 is in excess.

Convert 0.748 moles of SO2 to grams. O.748moles*(64.1g/mole) = 47.9 grams SO2.

To convert this into a volume (liters), we need to know the temperature and pressure. At STP (Standard Temperature and Pressure) is conversion is easy. Every gas occupies 22.4/mole of that gas at STP. But we are not told the exact temperature, other than “room” temperature. Being lazy, I’ll conveniently assume they are doing this experiment outdoors on the arctic tundra of Alaska, near the Bering Sea, during a warm day of 0 degrees C. That is standard temperature, and a pressure of 1 atm is not uncommon near the ocean at sea level. So, in the absence of actual numbers, we will assume we’re at STP. Therefore,

We use the moles value to state that 0.748 moles of SO2 gives us 16.7 liters (0.748 moles * (22.4 liters/mole) = 16.7 liters. We only have 2 sig figs (from 24g sulfur) so the answer is 17 liters.

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2) Hydrogen peroxide, H2O2, decomposes according to the equation:

2H2O2(l)-> 2H2O(l)+O2(g)

calculate the volume of oxygen, measured at room conditions, that can be obtained from the decomposition of 17g of hydrogen peroxide

The equation is balanced. Start by calculating the moles of hydrogen peroxide.

17g / (32g/mole) = 0.5 moles H2O2

We get 2 moles of O2 for every mole of H2O2 decomposed. Assume all the H2O2 decomposes, then we get 0.5 moles H2O2 * 2 = 1.0 moles O2. [Convenient].

As before, we are not given values for the gas conditions, so we’ll assume STP again. Make up your own scenario this time. In a lab hood, maybe?

Volume of O2 = 1.0 moles O2 * (22.4 liters/mole) = 22.4 liters of O2.

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3) a) Find mass of copper with equation 2Cu0(s)+C(s)->2Cu(s)+CO2(g) given 80g of CuO

The equation is balanced. We start with 80g of CuO, which has a molar mass of 79.5 g/mole. That converts to 1.006 moles of CuO. Make that 1.0 mole (2 sig figs). Assume an excess of C, carbon. We get 1 mole of copper for every 1 mole of CuO reacted. Therefore, we will get 1.0 moles of Cu. Copper’s molar mass is 63.5 g/mole. 1.0 mole of Cu is thus equal to 63.5 grams of Cu. We need to round to 2 sig figs: 64 grams of Cu.

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b) Find Volume of carbon dioxide given the same equation as (a) and given the same amount as (a)

The equation says we get 1 mole of CO2 per 2 moles of CuO. As before, 80 grams of CuO is 1.0 moles of CuO. Theat means we only get 0.5 moles of CO2. The molar mass of CO2 is 44 grams/mole CO2. Therefore, we get 0.5 moles * 44g/mole = 22 grams of CO2.

Assuming STP (same story), the volume of CO2 is 0.5moles CO2 * 22.4 liters/mole = 11.2, or 11 liters of CO2.

2ZnS(s)+302(g)->2ZnO(s)+2SO2(g)

given 12dm3 of SO2

c) Find Mass of zinc sulfide

Wow. Who wants to fuss around with dm^3 at this time of the morning? Oh well, let’s have at it. Let’s convert it into a unit we love and feel comfortable with -** liters**.

We know that:

1 liter = 1,000cm^3

And, 1 dm = 0.1 meter, or: 1 meter = 10 dm

even, 1 cm = 0.01 meter,

Let’s convert the 1dm^3 into cm^3: 1dm^3 * (1meter/10 dm)^3*(1 cm/0.01 meter)^3 = ????

Before I multiply this out, cross off the units: The first term eliminates the dm^3 and replaces it with meter^3. The second replaces meter^3 with cm^3. AHA! I do not know what this means, but the units cancel, and so I am **happy**.

Multiplying this out, we get 1000cm^3. Since we start with 1 dm^3, my conversion factor is 1000cm^3/dm^3. I also know that 1 liter = 1,000cm^3. Therefore, I can express the 12 dm^3 either as:

· cm^3 :: 12 dm^3 * 1000cm^3/dm^3 = 12,000cm^3

· liters :: 12,000cm^3 * (1 liter/1000 cm^3) =12 liters

Why didn’t they say so to begin with??

We have to assume STP, although I’m getting nervous by now. 12 liters of SO2 at STP gives us the following number of moles:

12 liters SO2/(22.4 liters/mole) = 0.54 moles.

Sorry, I’ve got to leave and cannot finish the problem, You should be able to complete the last questions following the processes we’ve gone through already. I’ll try to get back to it later today.

Bob