Answer:
ZnSO4 + 2LiNO3 → Zn(NO3)2 + Li2SO4
Explanation:
There's many resources on web that can assist you with this concept:
https://en.intl.chemicalaid.com/tools/equationbalancer.php
https://www.webqc.org/balance.php
Answer:
B: The sulfuric acid is not consumed or react with the reactant.
It has 8 bonding electrons
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!
Answer:
6.791
Explanation:
For proper significant figures with addition, you would use the significant figures of the number with lowest decimal place. 6.298 goes to the 10⁻³ place. 0.492712 goes to the 10⁻⁶ place. You will go out to the 10⁻³ place.
6.298 + 0.492712 = 6.790712 ≈ 6.791