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$\text{Ta có $2$ PTHH sau:}$

$1/$ $2Na + H_{2}O → 2NaOH + H_{2}$

$2/$ $H_{2} + CuO$ $\xrightarrow[]{t^o}$ $Cu + H_{2}O$

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$\text{Ta có:}$ $n_{CuO}$ $=\frac{40}{80}=0,5(mol)$

$\text{Theo PTHH $1$:}$ $n_{H_{2}}=$ $n_{CuO}$ $=0,5(mol)$

$\text{Theo PTHH $2$:}$ $n_{NaOH}=$ $2n_{H_{2}}$ $=0,5.2=1(mol)$

$⇒$ $m_{NaOH}=n.M=1.40=40(g)$

$C$`%_{NaOH}`$=$$\frac{40}{160}.100$`%`$=25$`%`

Đáp án:kk

Giải thích các bước giải:

Ta có duy nhất 22 PTHH sau:1) 2Na+2H2O2Na+2H2O →→ 2NaOH+H22NaOH+H22) H2+CuOH2+CuO to→→to Cu+H2OCu+H2O_____________________________________________________Có nCuO=4080=0,5 molnCuO=4080=0,5 mol

Dựa vào PTHH 2) nH2=nCuO=0,5molnH2=nCuO=0,5molDựa vào PTHH

1) nNaOH=2nH2=0,5.2=1molnNaOH=2nH2=0,5.

2=1mol

Vậy mNaOH=n.M=1.40=40gm

NaOH=n.M=1.40=40g→C%NaOH=40160.100%=25%