Through December 21st, KFC is offering a free Chicken Sandwich (classic or spicy) with any $5 or more purchase when you use promo code KFCFREE11 at checkout. The offer will be auto-applied in cart and you can choose free in-restaurant pickup to save on additional fees.
Have in mind that you need to make a minimum purchase of $5 in order for the deal to apply. There’s a limit of one per customer and offer cannot be combined with any other digital offers in the same transaction. Donât miss out, and order at KFC now!
KFC Original Honey BBQ
KFC fans, attention! Visit your nearest KFC, where you can enjoy the Original Honey BBQ again! You can combine this sweet and smoky flavor meal with chicken, tenders, and more, or grab a single drum for just $2! Please note that prices may vary by location. See all the options here, and enjoy your favorite KFC Original Honey BBQ meal!
KFC Hot & Spicy Wings 20-Piece for Just $20
Visit KFC where Rewards Members can score 20 Hot & Spicy Wings and four dipping sauces of their choice for just $20. Note that this offer is available only through the KFC app. Donât miss out, download the KFC app now (iOS or Android) and order at KFC now!
KFC 8-Piece Fried Chicken Bucket
On Tuesdays, when you purchase online or through KFC app, select KFC Restaurants are offering an 8-Piece Fried Chicken Bucket (Original, Extra Crispy, or Kentucky Grilled) for just $10. (Android or iOS). Orders must be placed on KFC.com or the KFC app and thereâs a limit of 1 per transaction.
You can redeem this offer by clicking the banner on the menu page. Donât miss out, order at KFC now!
Browse the latest restaurant deals here!Â
Awesome there goes the diet đ
Great Deal !!! Thanks, May Try it for dinner tonight
Most welcome Casandra!
Oh wow đŽ
But the restaurant near to me doesnât offer this promo
I’m sorry to hear that!
it’s true the Free stuff
Hi! We share a lot of deals alongside free samples. Here are the latest freebies for you to browse: https://www.freestufffinder.com/free-stuff/ Let me know if I can help you out somehow k?