can anyone explain to me how to balance a chemical equation? im having trouble in understanding how i keep getting it wrong.

2 answers:

6 0

For example we are going to use this unbalanced chemical reaction:
H₂ + O₂ → H₂O.
First, calculate number of atoms (hydrogens and oxygens) on left and right. There is two oxygen and two hydrogen on left and two hydrogen and one oxygen on right.
You can not change molecular formula of compound, only you can put coefficient in fron of compound to balance reaction.
Put 2 in front water to balance oxygen (now you have two oxygens on left and right). But now you have four hydrogens on right, so you must put 2 in fron hydrogen on the left.
2H₂ + O₂ → 2H₂O.

Elan Coil [88]3 years ago

5 0

<span>To balance a chemical equation, one has to make the total number of molecules of each element equal on both sides of the equation.
For example,
in the reaction of sodium (Na) and chlorine (Cl2), we get sodium choride (NaCl)
Na + Cl2--->NaCl
Now one chlorine molecule contains 2 atoms of Cl and one Na molecule contains 1 atom of Na
To balance, we have to take 2 Na molecules,
2Na + Cl2 ---> NaCl
Balancing both sides we get
2Na + Cl2 = 2NaCl
Here both sides of the chemical equation contains equal number of molecules of each element, so it is a balanced equation.</span>

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When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H

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$Y=58.15\%$

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Hello,

For the given chemical reaction:

$Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)$

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

$n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3$

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

$n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3$

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

$m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3$