a)
Ta có:
\({n_{Zn}} = 2,5{\text{ mol}} \to {{\text{m}}_{Zn}} = 2,5.65 = 162,5{\text{ gam}}\)
\({n_{CuS{O_4}}} = 0,5{\text{ mol}} \to {{\text{m}}_{CuS{O_4}}} = 0,5.(64 + 96) = 80{\text{ gam}}\)
b)
Ta có:
\({n_{C{O_2}}} = 2,5{\text{ mol}} \to {{\text{m}}_{C{O_2}}} = 2,5.(12 + 16.2) = 110{\text{ gam}}\)
c)
Ta có:
\({n_{C{O_2}}} = \frac{{11,2}}{{22,4}} = 0,5{\text{ mol}}\)
\( \to {m_{C{O_2}}} = 0,5.(12 + 16.2) = 22{\text{ gam}}\)
d)
Ta có:
\({n_{NaOH}} = \frac{{{{12.10}^{23}}}}{{{{6,023.10}^{23}}}} \approx 2{\text{ mol}}\)
\( \to {m_{NaOH}} = 2.(23 + 16 + 1) = 80{\text{ gam}}\)