3.5- Calculating an average atomic mass is like doing a weighted average to find your grade. You take the relative abundance (%) times the atomic masses of the respective isotopes, and add. So, 34.968 x 0.7553 + 36.956 x 0.2447 = answer. The answer should match the number on the periodic table.

3.65- Multiply by the mole ratio using train tracks. 3.60 mol CO x 2 mol CO2 /2 mol CO. The question was asking for moles of CO2 so you're done.

3.77- grams -> moles -> moles-> grams. Convert 1E5 to grams, then to moles of (NH4)2SO4. Then multiply by the mole ratio (moles NH3/ moles (NH4)2SO4). Then covert moles of NH3 to grams, then to kg. It can all be done on train tracks. See example 3:13. It is similar but doesn't have the gram<-> kg conversions.

3.83- Do the stoichiometry problem twice. Multiply each amount by the respective mole ratio. So, 0.886 mol NO x 2 NO2/2NO. gives 0.886 moles IF NO is limiting. Now, 0.503 mol O2 x 2NO2/1O2 makes 1.006 moles IF O2 is limiting. Thus, the limiting reactant is NO and the maximum amount of NO2 is 0.886 moles. See the top half of pg 104 for an example.

3.89- To calculate percent yield you need to a stoichiometry problem to find out the theoretical number of kg of HF. So, grams ->moles -> moles -> grams again. Then, once you have a theoretical number of kg of HF, find percent yield. Take the actual yield (2.86 kg) and divide by theoretical. Lastly, turn it into a percent by multiplying by 100. Example 3:16 can help.

3.65- Multiply by the mole ratio using train tracks. 3.60 mol CO x 2 mol CO2 /2 mol CO. The question was asking for moles of CO2 so you're done.

3.77- grams -> moles -> moles-> grams. Convert 1E5 to grams, then to moles of (NH4)2SO4. Then multiply by the mole ratio (moles NH3/ moles (NH4)2SO4). Then covert moles of NH3 to grams, then to kg. It can all be done on train tracks. See example 3:13. It is similar but doesn't have the gram<-> kg conversions.

3.83- Do the stoichiometry problem twice. Multiply each amount by the respective mole ratio. So, 0.886 mol NO x 2 NO2/2NO. gives 0.886 moles IF NO is limiting. Now, 0.503 mol O2 x 2NO2/1O2 makes 1.006 moles IF O2 is limiting. Thus, the limiting reactant is NO and the maximum amount of NO2 is 0.886 moles. See the top half of pg 104 for an example.

3.89- To calculate percent yield you need to a stoichiometry problem to find out the theoretical number of kg of HF. So, grams ->moles -> moles -> grams again. Then, once you have a theoretical number of kg of HF, find percent yield. Take the actual yield (2.86 kg) and divide by theoretical. Lastly, turn it into a percent by multiplying by 100. Example 3:16 can help.