Question

write the balanced chemical equation for the following reaction : a. sodium reacts with oxygen b. aluminum reacts with dilute hydrochloric acid​

Answer 1

$\huge{\star}{\underline{\boxed{\red{\sf{Answer :}}}}}{\star}$

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$\large{\sf{(a) \: Na \: + \: O \to \: Na_{2}O}}$

$\large{\sf{(b) \: 2Al \: + \: 6HCl \: \to \: 2AlCl_{3} \: + \: 3H_{2}}}$

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Additional Information :-

For Balancing a equation we need to follow the given steps :

1. Reactants are written on left side and Product to the right side.
2. We put a arrow between Reactants and Products.
3. Every atom after and before the reaction should be equal .

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Answer 2

Here is your solution⬇️

$\bold{\red{\boxed{\boxed{\huge\underline{1.}}}}}$

$\bold{\red{sodium\: react \:with\: oxygen}}$

Na+o2----------->Na20

$\underline{\purple{\bold{Now\:we \:balance\: reaction. }}}$

2Na+o2----------->Na20

This is balance chemical reactions .

$\bold{\purple{\boxed{\boxed{\huge\underline{2.}}}}}$

$\bold{\red{Aluminium\: react \:with\: dilute}}$

$\bold{\red{Hydrochloric\: acid.}}$

Al + Hcl -------> Alcl3+H2

$\underline{\purple{\bold{Now\:we\:balance \:reaction.}}}$

2Al + 6Hcl -------> 2Alcl3+3H2

This is balance chemical reactions.