Đáp án:
$ {m_{CaC{O_3}}} = 56\,\,gam $
Giải thích các bước giải:
$ Gly - Gly: N{H_2} - C{H_2} - CONH - C{H_2} - COOH$
${n_{Gly - Gly}} = 0,14\,\,mol$
$ \to {n_{C{O_2}}} = 4.{n_{Gly - Gly}} = 4.0,14 = 0,56\,\,mol $
$ C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O $
$ 0,56 \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,56\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol $
$ \to {m_{CaC{O_3}}} = 0,56.100 = 56\,\,gam $