#### Answer

$6.55g\ CaCO_{3}$

#### Work Step by Step

$3.67g\ CaO\times\frac{1mol\ CaO}{56.077g\ CaO}\times\frac{1mol\ CaCO_{3}}{1mol\ CaO}\times\frac{100.087g\ CaCO_{3}}{1mol\ CaCO_{3}} = 6.55g\ CaCO_{3}$
We divide the mass in grams of the reactant by its molar mass to get the number of moles of reactant. we then multiply this by the ratio obtained from the balanced chemical equation with the reactant on the bottom and the product on the top. We now have the number of moles of product. We multiply this by the molar mass of the product to get our result.