You can chain the replace calls:

mylist = mylist.replace(`some stuff`, 'things').replace(`more stuff`, 'many things');

If you want to store the pairs in an object for scalability, you can loop through the object's entries (with Object.entries) and replace all occurences of each key with its value:

const obj = {
  "some stuff": "things",
  "more stuff": "many things"
}

var str = "<div>some stuff</div>A lot of code in between<span>more stuff</span>"

Object.entries(obj).forEach((k,v) => str = str.replace(k,v))

console.log(str)

You can first create a changes object which contain key-value of replace-replacement and then iterate over the key-value and replace the key with the value

This is better because if you have multiple replacements then It will work every time if you just add more replacer - replacement in the object itself.

You don't have to add extra replace statement, just add key-value in the changes and you are good to go.

let mylist = `<div>some stuff</div>A lot of code in between<span>more stuff</span>`;

const changes = {
  "some stuff": "things",
  "more stuff": "many things",
};

mylist = Object.entries(changes).reduce((str, [k, v]) => str.replace(k, v), mylist);
console.log(mylist)