Question

Fe + Cl2 —> FeCl3.
Balance the equation .
Please, give the answer showing how it has came. ​

Answer 1

Answer:

See the attachment for balanced equation

Step-by-step explanation :

First see the equation.

You will notice it is not balanced one.

So, you need to balance it.

So, we can observe that in reactant side we have one atom of iron (Fe) and 2 atom of Cl and in the product side we have one atom of iron and 3 atom of chlorine so,

For making it a balanced equation we need to put 2 in the product side to get 6 atom of chlorine we put 2 here bcoz we get 6 and in reactant side we put 3 as co-officiant to get 6 atom of chlorine so, we are balancing chlorine which is done now, but in case of making chlorine balanced we get 2 atom of iron in the product side so, make it balanced we put 2 as co-officient of iron in the reactant side.

Hope you understand.

See the attachment

Answer 2

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Ans ➭ 2Fe + 3 Cl2 → 2FeCl3

steps ➭

$\: \sf \: lets \: wright \: down \: first \: how \: \\ \sf \: many \: atom \: we \: have \: on \: each \: side \:$

$\: \sf \: on \: reactant \: side \: \\ \sf \: (fe) = > \: 1 \\ \sf \: (cl) = > \: 2$

$\: \sf \: on \: product \: side \: \\ \sf \: (fe) = > \: 1 \: \\ \sf \: (cl) \: = > \: 3 \:$

$\: \sf \: so \: the \: (fe) \: is \: balanced \: \\ \sf \: but \: the \: (cl) \: is \: not \: \: \: \: \: \: \: \: \: \:$

$\: \sf \: we \: have \: an \: odd \: number \: as \: (3) \: on \: \\ \sf product \: side \: lets \: get \: it \: even \: by \: \: \: \: \: \\ \sf \: by \: multiplying \: it \: by \: (2) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \: so \: 3 \times 2 = 6 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\: \sf \: so = > \: Fe + Cl2 —> 2FeCl3.$

$\: \sf \: we \: also \: have \: to \: make \: (fe) \: 1 \times 2 \: to \: \\ \sf \: make \: sure \: our \: counting \: is \: correct \: \: \: \\ \sf \: lets \: multiply \: it \: 1 \times 2 = 2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf \: over \: on \: a \: reactant \: side \: we \: have \: (fe \: = 1 \: ) \\ \sf \: but \: we \: need \: to \: have \: (2) \: so \: we \: will \: just \: \: \: \: \: \: \\ \sf \: put \: the \: coefficient \: of \: (2) \: and \: then \: the \: (1 \times 2 = 2)$

$\sf \: so \: = > \: 2Fe + Cl2 → 2FeCl3$

$\sf \: so \: (fe) \: is \: now \: balanced \: on \: both \: the \: sides \:$

$\sf \: and \: now \: we \: need \: to \: have \: 6 \: (cl)atoms \\ \sf \: on \: both \: the \: sides \: so \: but \: i \: \\ \sf \: have \: 2 \: on \: the \: reactant \: side \: \\ \sf \: so \: lets \: multiply \:( 2 \times 3 = 6)$

$\sf \: so = > (2)Fe + (3)Cl2 → (2)FeCl3$

$\sf \: and \: our \: equation \: is \: balanced$

hope it was helpful to you