NaN₃ is the chemical formula for Sodium Azide
<span>atomic weights: Al = 26.98, Cl = 35.45
In this reaction; 2Al = 53.96 and 3Cl2 = 212.7
Ratio of Al:Cl = 53.96/212.7 = 0.2537 that is approximately four times the mass Cl is needed.
Step 2:
(a) Ratio of Al:Cl = 2.70/4.05 = 0.6667
since the ratio is greater than 0.2537 the divisor which is Cl is not big enough to give a smaller ratio equal to 0.2537.
so Cl is limiting
(b)since Cl is the limiting reactant 4.05g will be used to determine the mass of AlCl3 that can be produced.
From Step 1:
212.7g of Cl will produce 266.66g AlCl3
212.7g = 266.66g
4.05g = x
x = 5.08g of AlCl3 can be produced
(c)
Al:Cl = 0.2537
Al:Cl = Al:4.05 = 0.2537
mass of Al used in reaction = 4.05 x 0.2537 = 1.027g
Excess reactant = 2.70 - 1.027 = 1.67g
King Leo · 9 years ago</span>
Answer:
2Na + Br2 = 2NaBr
Explanation:
In order to balance a chemical equation you make the make sure both sides have the same number of atoms on each side, you do this by multiplying on both sides as if it was a algebraic equation.
Na+ Br2 = NaBr
Na × 2 = Na2
Na × 2 = Na2
Br × 2 = Br2
2Na + Br2 = 2NaBr
Hope this helps.
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