The mass of gas A is 11.56. i got this answer by 0.68 multiplied by 17. because in this problem the key word is times.
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!
Use the periodic table , it should be the long decimal if u look it up
Answer:
25.1 g is the mass of chlorine in the sample
Explanation:
P . V = n . R . T
We apply the Ideal Gases Law to solve the excersise.
We need to convert the T°C to T°K → 37°C + 273 = 310K
We replace data: 3 atm . 3L = n . 0.082 L.atm /mol.K . 310K
9 atm.L / (0.082 L.atm /mol.K . 310K) = n → 0.354 moles
We convert the moles to mass, to reach the answer
0.345 mol . 70.9g / 1mol = 25.1 g