$\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{}\overline{){\mathbf{cm}}^{\mathbf{3}}\mathbf{}\mathbf{Ba}}\mathbf{}\mathbf{\times}\mathbf{}\frac{\mathbf{3}\mathbf{.}\mathbf{51}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{Ba}}}{\mathbf{1}\mathbf{}\overline{){\mathbf{cm}}^{\mathbf{3}}\mathbf{}\mathbf{Ba}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{Ba}}{\mathbf{137}\mathbf{.}\mathbf{327}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{Ba}}}$ **= 0.06390 mol Ba**

$\mathbf{1}\mathbf{.}\mathbf{75}\mathbf{}\overline{){\mathbf{cm}}^{\mathbf{3}}\mathbf{}\mathbf{S}}\mathbf{}\mathbf{\times}\mathbf{}\frac{\mathbf{2}\mathbf{.}\mathbf{07}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{S}}}{\mathbf{1}\mathbf{}\overline{){\mathbf{cm}}^{\mathbf{3}}\mathbf{}\mathbf{S}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{S}}{\mathbf{32}\mathbf{.}\mathbf{07}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{S}}}$ **= 0.1130 mol S**

When barium (Ba) reacts with sulfur (S) to form barium sulfide (BaS), each Ba atom reacts with an S atom. If 2.50 cm^{3 }of Ba reacts with 1.75 cm^{3} of S, are there enough Ba atoms to react with the S atoms (*d* of Ba = 3.51 g/cm^{3}; *d* of S = 2.07 g/cm^{3})?

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