Question

In research on the synthesis of superconductors, the following reaction is studied. What amount (in moles) of BaO is needed to react with 5.0g of CaO?

Tl₂O₃(l) + 2 BaO(s) + 3 CaO(s) + 4 CuO(s) $\rightarrow$ Tl₂Ba₂Ca₃Cu₄O₁₂(s)

Answer 1

Tl₂O₃(l) + 2 BaO(s) + 3 CaO(s) + 4 CuO(s) -----> Tl₂Ba₂Ca₃Cu₄O₁₂(s)

From the balanced equation:

3 moles of CaO requires -----> 2 moles of BaO

moles of CaO = mass/molarmass      ( CaO = 56 g/mol)

                        = 5.0/56

                        = 0.0893 moles

3 moles of CaO requires -----> 2 moles of BaO

0.0893 moles of CaO requires ----> (2*0.0893)/3

                                                          = 0.0595 moles of BaO

moles of BaO needed = 0.0595 moles