2К + 2Н2О = 2КОН + Н2

n(K)=m/M=7,8/39=0,2(моль)  

n(H2)=0,1 моль  

V(H2)=n*Vm=0.1*22.4=2.24дм3

дано

m(K) = 7.8 g

η(H2) = 70%

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Vпракт (H2)-?

2K+HOH-->2KOH+H2

M(K) = 39 g/mol

n(K) = m(K) / M(K) = 7.8 / 39 =0.2 mol

2n(K) = n(H2)

n(H2) = 0.2 / 2 = 0.1 mol

Vтеор(H2) = n(H2) * Vm = 0.1*22.4 = 2.24 L

Vпракт (H2) = V(H2)*η(H2) / 100% = 2.24 * 70% / 100% = 1.568L

ответ 1.568 л