Iwill give a brainliest answer to 1st person to answer !
i have a super important chem test tomorrow! show all work on both problems and do both! it would be super super to me if you explained why those are the answers.

The whole idea being limiting reagents:
when a chemical reaction happens, the reactants get used up. you can only have a chemical reaction when you have all the reactants present
one reactant can get completely used up before the other reactant; that would be the limiting reactant.



(1)

we need a balanced chemical equation first before we can determine the maximum amount.

potassium chloride:  

barium sulfate:  

these combine together to make a double replacement reaction. the compounds swap their ions. the double replacement reaction between potassium chloride and barium sulfate would look like

   

note that we have  as an ion, so when it swaps, the potassium needs a subscript of 2. and we also have  as an ion, so when it swaps, the chlorine needs a subscript as well.

balancing the above equation:

   

now, if we want to determine the limiting reactant, we need to use our information. first we can calculate the moles of  producible with 100.0 grams of potassium chloride.

molar mass of potassium chloride is 74.6 g/mol, so converting grams to moles of kcl

   

i leave it unfinished because we have one more step.

notice from the balanced chemical equation we have a ratio between , what we have, and  , which is what we want. this ratio is

   

so we can multiply the moles of kcl by this ratio to get the number of moles of    formed from  . (notice how in the ratio, i put kcl in the numerator so the units properly cancel.)

   

so with 100.0 grams of potassium chloride, we can form 0.670 moles of barium chloride. when 100.0 g of potassium chloride is used up in the reaction, we form 0.670 moles of barium chloride. 

now figure out the number of moles of    formed from  .

the molar mass of    233.4 g/mol.
the ratio between barium chloride and barium sulfate, from the balanced chemical equation, is 1: 1

   

we would be able to produce 1.71 moles of bacl2 with 400.0g of baso4.
but we are only able to produce 0.670 moles of bacl2 with 100.0g of kcl.

kcl will get used up first, producing 0.670 moles intotal. after that,  no reaction can happen because there will be no more kcl. we are only able to produce the amount that kcl can produce before it gets used up.

so the maximum amount of barium chloride producible is 0.670 moles.



(2)

standard combustion reaction:
some carbohydrogen plus oxygen gas gets us carbon dioxixide and water.

   

balancing this, we get

   

let us get the moles of carbon dioxide we can produce with 96.0 grams of c2h4.

molar mass of c2h4 is 28.0 g/mol.
the ratio between co2 and c2h4 2: 1 (2 co2 for 1 c2h4)

   

so we produce  6.85714 moles (i'm not rounding yet) of co2 with 96.0 grams of c2h4

with the 140.0 l of oxygen, we need to convert liters into moles.
since we are told that this is at stp, we know that there are 22.4 l/mol for a gas. calculating the moles of oxygen:

   

from bal equation, the ratio of co2 to o2 is 2: 3
calculating moles of co2 from moles of o2:

   

we produce less moles with oxygen, so we have to the number of co2 moles producible with oxygen because it'll get used up first.

molar mass of co2 is 44 g/mol, so converting

   

183 g is the is the max grams of co2