Consider the following reaction:
CO2(g) + 2NaOH(s) Na2CO3(s) + H2O(l)

(a) Identify the limiting reactant when 120.0 g CO2 and 185.0 g NaOH are allowed to react?

(b) How many grams of Na2CO3 can be produced?
____245.2 g___ (answer must be in 4 sig figs)

(c) If in lab only 195.0 g…

I’m assuming the equation is CO2(g) + 2NaOH(s) —> Na2CO3(s) + H2O(l)

a) 120.0g CO2 (1mole/44.0g CO2)(1m Na2CO3/1mole CO2) = 2.727 m Na2CO3

185.0g NaOH (1mole/40.0g NaOH)(1 m Na2CO3/2m NaOH) = 2.313 m Na2CO3

NaOH is the limiting reagent because it produces the least amount of Na2CO3 and thus there is only enough to make that much moles of Na2CO3.

b) using 2.313 moles of Na2CO3 as we solved from above.
2.313 moles Na2CO3 (106.0 g / 1 mole Na2CO3) = 245.2 g

c) % yield = 100 * actual mass / predicted mass
= 100 * 195.0 g / 245.2 g

I hope this helps! 🙂
Lemme know if you have any questions.

Really? Do you know what the right answer is then? I’m pretty sure 79.52% should be right because I just finished AP chem. If you have the right answer, I can probably figure out what was wrong/happened.

given off. The closer the flexibility stages are to the nucleus (draw of the – electron to the + center), the extra stable this is. so which you would be able to desire to fall from a a ways better orbit to a decrease orbit releases ability. Umm..assumption one: issues tend to prefer to pass to a extra stable state. 2…..negatively charged electron are attracted to the + nucleus yet that’s tide in with assumption one. . GL

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