Đáp án:
a) $20$ gam
b) $4,4$ gam
c)
$V_{CO} = 2,8$ lít
$V_{ NO_2} = 16,8$ lít
Giải thích các bước giải:
a)
Ta có:
\({m_{NaOH}} = {n_{NaOH}}.{M_{NaOH}} = 0,5.(23 + 16 + 1) = 20{\text{ gam}}\)
b)
Ta có:
\({n_{C{O_2}}} = \dfrac{{{V_{C{O_2}}}}}{{22,4}} = \dfrac{{2,24}}{{22,4}} = 0,1{\text{ mol}}\)
\( \to {m_{C{O_2}}} = {n_{C{O_2}}}.{M_{C{O_2}}} = 0,1.(12 + 16.2) = 4,4{\text{ gam}}\)
c)
\({V_{C{O_2}}} = {n_{C{O_2}}}.22,4 = 0,125.22,4 = 2,8{\text{ lít}}\)
\({V_{N{O_2}}} = {n_{N{O_2}}}.22,4 = 0,75.22,4 = 16,8{\text{ lít}}\)
(ghi chú: $Na = 23$)