Verified answer

BaCl2(aq) +K2CrO4(aq)->BaCrO4(s) + 2KCl (aq)

since there is much more reactivity in Ba, there will be the formation of precipitate, in this case BaCrO4

Bacl2 K2cro4

BaCl2(aq) +K2CrO4(aq)->BaCrO4(s) + 2KCl (aq)

http://www.webqc.org/balance.php