iron combines with oxygen to form rust. Given the chemical reaction, how many grams of 4Fe+3o2 would produce 10 g of rust.
The chemical reaction for the formation of rust is as follows: 4Fe + 3O2 –> 2Fe2O3. This indicates that, in order to produce 10 grams of rust, one needs 8 grams (4 x 2) of Iron and 6 grams (3 x 2) of Oxygen.
To calculate the number of moles of Iron and Oxygen required to make 10g of rust, we need to use their respective molecular weights – 56.07 g/mol for Iron and 32g/mol for oxygen. If you divide 8 g by 56.07 g/mol, it equals 0.14248 moles; similarly if you divide 6g by 32 g/mol it equals 0.1875 moles .
iron combines with oxygen to form rust. Given the chemical reaction, how many grams of 4Fe+3o2 would produce 10 g of rust.
Now that we know the number of moles required to form 10Gm Rust from Iron and Oxygen , let’s find out how many grams are needed in total: we multiply our mole values with the respective molecular weight : 0.14248 * 56 = 8 gram iron ; and 0.1875*32=6 gram oxygen; so in total 8+6=14grams are needed to produce 10grams or rust
In conclusion ,to produce 10 Grams of Rust from 4 Fe + 3 O2 would require 14 gram in total; this includes 8 Gram Iron and 6 Grams Oxygen
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