Gr 10 chem help uwu

[Poem]

Sulfur Dioxide is slowly oxidised in the atmosphere to form sulfur trioxide

I have to make a word + balanced equation for this
I put
SO2 + O2 -> SO3
???
Am I on the right track? IDK

Yup, just balance it like juliaxo said
Word equation would just be "sulfur dioxide reacts with oxygen to produce sulfur trioxide",

[Maki]

What do you mean by 'make a word'?
If your equation is SO2 + O2 -> SO3 then to balance it you'd just make it 2SO2 + O2 -> 2SO3

how did u get that?? how did u even know to put those coefficients. still confused
when i see stuff like this for example
reactants:
S- 3
products:
S- 4
i start freaking out bc there is no easy coefficient to put,,,

compared to this-
reactants:
S-2
products:
S- 4
bc for that ^,
i can just put a coefficient of 2 on the reactant side
its like easy multiplication
ja feel?

[Jules]

how did u get that?? how did u even know to put those coefficients. still confused

Because whatever is on the products side has to be equal to what is on the reactants side. So, your original equation was SO2 + O2 -> SO3, and you have 1 sulfur and 4 oxygen on the reactant side, but 1 sulfur and 3 oxygen on the products side. So by adding the coefficients and making it 2SO2 + O2 -> 2SO3, you now have 2 sulfur and 6 oxygen on the reactants side and the products side, so the equation is balanced.

when i see stuff like this for example
reactants:
S- 3
products:
S- 4
i start freaking out bc there is no easy coefficient to put,,,

compared to this-
reactants:
S-2
products:
S- 4
bc for that ^,
i can just put a coefficient of 2 on the reactant side
its like easy multiplication
ja feel?

If I had numbers like S 3 and 4 I just multiply by whatever will make both the same number. It takes a while to get used to all of it, but my teacher would give us really complicated combustion reactions to balance where your ending coefficients of the balanced equations are numbers like 17 and 51 and they take like 20 minutes to balance, it really is just guessing and checking half the time.

[Maki]

Because whatever is on the products side has to be equal to what is on the reactants side. So, your original equation was SO2 + O2 -> SO3, and you have 1 sulfur and 4 oxygen on the reactant side, but 1 sulfur and 3 oxygen on the products side. So by adding the coefficients and making it 2SO2 + O2 -> 2SO3, you now have 2 sulfur and 6 oxygen on the reactants side and the products side, so the equation is balanced.



If I had numbers like S 3 and 4 I just multiply by whatever will make both the same number. It takes a while to get used to all of it, but my teacher would give us really complicated combustion reactions to balance where your ending coefficients of the balanced equations are numbers like 17 and 51 and they take like 20 minutes to balance, it really is just guessing and checking half the time.

Ahh I see!
thanks

could you explain balancing
HCl + Ca(OH)2 -> CaCl2 + H2O?
the last thing i got before raging in a fit was
4HCl + Ca(OH)2 -> 5CaCl + 3H2O lol

[Jules]

Ahh I see!
thanks

could you explain balancing
HCl + Ca(OH)2 -> CaCl2 + H2O?
the last thing i got before raging in a fit was
4HCl + Ca(OH)2 -> 5CaCl + 3H2O lol

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This is how I balance, I make REP tables (reactants, elements, and products) so it is easier to see what you gave on each side and what you need