10.0 g C2H2 x 1 mol CH2 x 1 mol CaC) = .384 mol CaCy 1 mol C2H2
А
3. What conversion factor belongs in the box marked A if you are trying to
determine the number of moles of CaC, needed to prepare 10.0 g C, H,? CaC, +
2H,0 → 12H2 + Ca(OH)2
O 1 mol Caca
26.049 C, H,
O 1 mol CH
10g Ca(OH)2