How many grams of caco3 are needed to react with 15.2g of hcl​

Balanced equation:

CaCO3 (s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)

1mol CaCO3 will react with 2 mol HCl

Mol HCl in 32.8mL of 1.25M solution = 32.8mL / 1000mL/L * 1.25mol /L = 0.041 mol HCl

This will react with 0.041/2 = 0.0205 mol CaCO3

Molar mass CaCO3 = 100g/mol

Mass of CaCO3 reacted = 0.0205mol *100g/mol = 2.05g CaCO3.

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !