Hope you've still got time to look this over before your test. I don't have the time or will to answer each question, so I'll just show you how they're done and provide an example.
For each of the following unbalanced equations, calculate how many milligrams of each product would be produced by complete reactions of 10.0 mg of the reactant indicated in italics. Indicate clearly the mole ratio used for the conversion.
a. FeSO4 (aq) + K2CO3 -> FeCO3(s) + K2SO4(aq)
b. Cr(s) + SnCl4(l) -> CrCl3(s) + Sn(s)
c. Fe(s) + S8(s) -> Fe2S3(s)
d. Ag(s) + HNO3(aq) + H20(l) + NO(g)
To solve these problems, first you need to balance the equation so that the moles of each element on the left equal the number of moles of that element on the right. So let's look at
c as an example. The balanced equation would be:
16Fe(s) + 3S8(s) -> 8Fe2S3(s)
Next, you use the information you're given to calculate whatever the question asks for. The question says that you have to find the amount of product formed if given 10 mg of the reactant in italics; since you didn't italicize anything, I'll just pick Fe for this example. Say you're given 10 mg (.01 g) of Fe for this reaction. First calculate how many moles that is. The molar mass of Fe is 56g. So you have
(0.01g Fe)/(56 g/mol Fe)=0.00018 mols Fe. Assuming that the sulfur (S8) is present in excess, the Fe will react completely to produce the product.
This is where the molar ratios come in. For every 16 mols of Fe in the reactants, you have 8 moles of Fe2S3 produced, so if we have 0.00018 mol Fe, then we will get
(0.00018 mol Fe)*[(8 mol Fe2S3)/(16 mol Fe)], which can be written more simply as
(0.00018)*(1/2), or
0.00009 mol Fe2S3.
Now that we know the number of mols of product, we need to figure out the mass of the product. To do this, simply multiply the number of moles (in this case,
0.00009) by the molar mass (for Fe2S3, it's 208):
0.00009 mol Fe2S3*208 g/mol Fe2S3 = 0.01872 g Fe2S3, or
18.72 mg Fe2S3.
If you have multiple products, use the exact same process; you just need to figure out how many moles of each product are produced (based on the moles of reactant you're given) and then multiply by that product's molar mass.
Small quantities of oxygen gas can be generated in the lab by the decomposition of hydrogen peroxide. The unbalanced equation for the reaction is
H202(aq) -> H20 + 02(g)
Calculate the mass of oxygen produced when 10.00g of hydrogen peroxide decomposes.
Use the exact same method as above.
Ammonium nitrate has been used as a high explosive because it is unstable and decomposes into several gaseous substances. The rapid expansion of the gaseous substances produces the explosive force.
NH4NO3(s) -> N2(g) + 02(g) +H20(g)
Calculate the mass of each product gas if 1.25 g of ammonium nitrate reacts.
Same process as above.
Elemental fluorine and chlorine gases are very reactive. For example, they react with each other to form chlorine monofluoride.
Cl2(g) + F2(g) -> 2ClF(g)
Calculate the mass of chlorine gas required to produce 5.00 + 10^-3g of chlorine monofluoride given an excess of fluorine gas.
Same process, but this time in the opposite direction. You're given the amount of product. Figure out how many moles that is; then figure out how many moles of Cl2 is needed to make that (hint: for each mole of Cl2, two moles of ClF are produced. You need to use this fact to multiply by the right molar ratio). Once you've determined the moles of Cl2, just figure out the mass of Cl2 by multiplying by the molar mass of Cl2.
Both propane (C3H8) and butane (C4H10) react with oxygen gas to form carbon dioxide and water. If you have equal masses of each, which will require a greater mass of oxygen to react?
These are examples of a combustion reaction, so you can easily figure out the balanced equation for each.
For any alkane/hydrocarbon combustion reaction, you have the following equation (where x and y are the subscripts for carbon and hydrogen):
CxHy + ?O2 --> ?CO2 + ?H2O
For propane:
C3H8 + ?O2 --> ?CO2 + ?H2O
First use the fact that all the carbon is converted to CO2, so for each mole of carbon on the left, you'll have the same number of moles of CO2 on the right:
C3H8 + ?O2 --> 3CO2 + ?H2O
Then realize that all the hydrogen on the left goes into making H2O on the right, so all the moles of H on the left find their way into the H2O on the right, which allows us to determine this:
C3H8 + ?O2 --> 3CO2 + 4H2O
Now that we know the number of O on the right, we can fill in the last missing coefficient:
C3H8 + 5O2 --> 3CO2 + 4H2O
Using the same method for butane:
C4H10 + 6.5O2 --> 4CO2 + 5H2O
Now that you have balanced equations for each, you can come up with an answer. Question says that given the same mass of reactant hydrocarbon, which will use up more O2? Arbitrarily pick a mass; say, 100 g. Figure out how many moles of propane are in 100 g and how much O2 will be used (hint: for each mol of C3H8, you will use 5 moles of O2). Then do the same for butane (for each mol of C4H10, you will use 6.5 mols of O2). And there's your answer.