If 16.4 grams of copper (II) bromide react with 22.7 grams of sodium chloride, how many grams of sodium bromide are formed?
1 answer:
The amount of sodium bromide that would be formed from the reaction will be 7.5524 grams
<h3>Stoichiometric calculation</h3>Looking at the equation of the reaction:
The mole ratio of CuBr2 and NaCl is 1:2.
Mole of 16.4 grams of CuBr2 = 16.4/223.37
= 0.0734 moles
Mole of 22.7 grams of NaCl = 22.7/58.44
= 0.3884 moles
Equivalent mole of NaCl = 0.1468 moles
Thus, NaCl is in excess while CuBr2 is limiting.
Mole ratio of CuBr2 and NaBr = 1:1
Mass of 0.0734 mole NaBr = 0.0734 x 102.894
= 7.5524 grams
More on stoichiometric calculation can be found here: brainly.com/question/8062886
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Answer:
Explanation:
From the given information, since the molecular mass of the ion M+ is not given;
Let's assume M+ = 58.0423
So, by applying the 13th rule;
we will need to divide the mass by 13, after dividing it;
The quotient n = no. of carbon; &
The addition of the quotient (n) with the remainder r = no. of hydrogen.
So;
So;
From the given information; we have oxygen present, so since the mass of oxygen = 16, we put oxygen in the molecular formula by removing . Also, since the mass is an even number then Nitrogen is 0.
So, we have: