$n_{H_2SO_4}=0,1a$

$n_{NaOH}=0,15.1,5=0,225 mol$

$n_{Al}=0,025 mol$

- TH1: dư NaOH

$2NaOH+2Al+2H_2O\to 2NaAlO_2+3H_2$

$\Rightarrow n_{NaOH\text{dư, ban đầu}}=0,025.2=0,05 mol$

$2NaOH+H_2SO_4\to Na_2SO_4+H_2O$

$n_{NaOH\text{pứ}}=0,225-0,05=0,175 mol$

$\Rightarrow n_{H_2SO_4}=n_{Na_2SO_4}=0,0875$

$\Rightarrow 0,1a=0,0875$

$\Leftrightarrow a=0,875(M)$

Ở phần 2: $n_{Na_2SO_4}=0,0875:2=0,04375; n_{NaOH}=0,025 mol$

$\Rightarrow m_{\text{rắn}}=7,2125g$

- TH2: dư $H_2SO_4$

$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$

$\Rightarrow n_{H_2SO_4\text{dư, ban đầu}}=1,5.0,025.2=0,075 mol$

$n_{H_2SO_4\text{pứ}}=0,225:2=0,1125 mol$

$\Rightarrow 0,1a-0,1125=0,075$

$\Leftrightarrow a=1,875(M)$

$n_{Na_2SO_4}=0,5n_{NaOH}=0,1125 mol$

Ở phần 2: $n_{Na_2SO_4}=0,05625 mol$

$\Rightarrow m_{\text{rắn}}=7,9875g$