The chemical reaction would be written as:
2Na + Cl2 = 2NaCl
Since we are given the amounts of the reactants available for reaction, we have to determine the limiting reactant. And use this amount to calculate for the theoretical yield.
55 g Na ( 1 mol / 22.99 g ) = 2.39 mol Na
67.2 f Cl2 ( 1 mol / 70.9 g ) = 0.95 mol Cl2
Therefore, the limiting reactant would be Cl2 since it is the one consumed completely.
0.95 Cl2 ( 1 mol NaCl / 1 mol Cl2) = 0.95 NaCl produced from the reaction
Classification ,
correct me if i’m wrong
We make use of ratios to solve this question.
moles NaOH = c · V = 0.1923 mmol/mL · 26.66 mL = 5.126718 mmol
moles H2SO4 = 5.126718 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 2.563359 mmol
Hence
[H2SO4]= n/V = 2.563359 mmol / 42.45 mL = 0.06039 M
The answer to this question is [H2SO4] = 0.06039 M
Answer:2 ml.
Explanation:Volume. 2.5 g/ml = 5g/Volume. Volume = 5g/(2.5 g/ml). = 2 ml.
