According to the law of conservation of mass, the amount of BARIUM present of the reactants is the same as the amount present in the products (the precipitate).
(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt
(10.0 g) / (0.0480 mol) = 208.3 g/mol
So it must have been BaCl2, because the molar mass of Barium is 137 which leave 71 grams left. Since Barium is a +2 charge, it means the atom next to it must be twice. Chlorine mass is 35, which twice is 71
When a fuel burns, it combines with oxygen int the air and changes into the substances water and carbon dioxide. That's how Flammability is a chemical property.
Global win I think brace it say around small areas
The best way to balance an equation is to balance one atom at a time.
You start with two Au atoms on the left, so you know the coefficient of Au on the right has to be 2. So at first we get,
Au2S3 + H2 --> 2Au + H2S
Then, notice you have 3 sulfur atoms on the left, so you need three on the right.
Our equation becomes
Au2S3 + H2 --> 2Au + 3H2S
Lastly, we now have six hydrogen atoms on the right, and only two on the left, so we assign a three to the H2 on the left
Au2S3 + 3H2 --> 2Au + 3H2S Is the balanced final equation.
