Moles=volume*concentration
=0.1*.83
=.083 Moles of HC2H3O2
Mole ratio between HC2H3O2 and CO2 is 1:1
This means .083 Moles of CO2
Mass =Moles*Rfm of CO2
=.083*(12+16+16)
=3.7grams
Answer:
- 13.56 g of sodium chloride are theoretically yielded.
- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.
- 0.50 g of sodium nitrate remain when the reaction stops.
- 92.9 % is the percent yield.
Explanation:
Hello!
In this case, according to the question, it is possible to set up the following chemical reaction:

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

Therefore, the leftover of sodium nitrate is:

Finally, the percent yield is computed via:

Best regards!
First, lets balance the reaction equation:
4Fe + 3O₂ → 2Fe₂O₃
It is visible form the equation that 4 moles of Fe require 3 moles of O₂
Molar ratio Fe/O₂ = 4/3 = 1.33
Molar ratio O₂/Fe = 3/4 = 0.75
Now, we check the molar ratios present:
Fe/O₂ = 6.8/8.9 = 0.76
O₂/Fe = 1.31
Thus, Iron is the limiting reactant because its ratio is not being fulfilled while the ratio of O₂ is surpassed.
A.electrons are shared between two different nuclei
Answer:
subscript is 3
Explanation:
the subscript is the number that is slightly lower than a # which in this case it's O
this indicates that there's 3 oxygen atoms