How to display result of a View in a specific block? I have a View displaying the content of a Custom Content Entity.
The first output I created is a Page display (accessible with a path and an option in a menu).
As expected, this is displayed in the 'Content' region of the default theme (see capture below).
I'd like to output the result in a full width region of the theme, i.e. without the content of 'Sidebar First' (User account menu, Search).
I duplicated the Page Display as a Block Display and placed it. But, of course, it appears on every page of the site :-(
So I created a single basic page (node/xx) and I configured the block to show only in this page. But I had to hide 'User account menu' and 'Search' on this page too.
Is there a more simple way to achieve this?
You are going about this the wrong way. Start again with a page display for your view. 

Now you want to remove the sidebar from that page. Two ways you might do that:


Disable all blocks in the sidebar when the path matches the one for your view. You'll need to configure each one separately with the path criteria.
Override the page template with one that does not have the sidebar code. The template will be named something like page--my-path.html.twig (See https://www.drupal.org/docs/8/theming/twig/twig-template-naming-conventions).


There are other approaches but these are probably the simplest to implement.