Go routine and how to use chan to get task' result with buffered chan P2

naturals := make(chan int,10)
squares := make(chan int)

go func() {
   for x := 0; ; x++ {
      fmt.Println("Pushing")
      naturals <- x
   }
}()

go func() {
   for {
      fmt.Println("Getting out x")
      x := <-naturals      

      squares <- x * x      

      time.Sleep(time.Second*2)
   }
}()

for {
   fmt.Println(<-squares)
}

naturals := make(chan int,10)
As you can see that we use a buffered chan int with size 10 for naturals
Then it can pushing to this chan maximum 10 element without waiting for getting out.