Gọi số mol K là a
\(m_{KOH}=\dfrac{50.12}{100}=6\left(g\right)\)
PTHH: 2K + 2H2O --> 2KOH + H2
______a---------------->a------>0,5a
mKOH = 6 + 56a (g)
mdd (sau pư) = 39a + 50 - 2.0,5a = 50 + 38a (g)
=> \(\dfrac{6+56a}{50+38a}.100\%=15\%\)
=> a = 0,03 (mol)
=> x = 0,03.39 = 1,17(g)