Question

Many metals react with oxygen gas to form the metal oxide. for example, calcium reacts as follows. 2 ca(s) + o2(g) → 2 cao(s) you wish to calculate the mass (g) of calcium oxide that can be prepared from 4.23 g of ca and 2.87 g of o2. (a) what amount (mol) of cao can be produced from the given mass of ca? webassign will check your answer for the correct number of significant figures. mol (b) what amount (mol) of cao can be produced from the given mass of o2? webassign will check your answer for the correct number of significant figures. mol (c) which is the limiting reactant? calcium oxygen (d) how many grams of cao can be produced? webassign will check your answer for the correct number of significant figures. g

Answer 1

(a) Assuming the amount of O2(g) is not limiting the reaction, a mass of 4.23g of Ca(s) will produce an equal mass of CaO(s), hence it will produce 4.23g of CaO(s). According to their respective molar masses, we have the following CaO molar mass : Molar mass of Ca + Molar mass of O = 40.1 + 16 = 56.1 g/mol 4.23g of Ca will then produce : 4.23 / 56.1 = 0,07540107 mol of CaO. (b) With the same reasonment as above, and assuming the amount of Ca is not limiting, we have : 2.87g of O2 will produce : 2.87 / 56.1 = 0,051158645 mol of CaO. (c) From (a) and (b) answers, we can conclude that the reactant that produces less mol of CaO is limiting the reaction. Hence following the given masses, O2 is the limiting reactant. (d) Knowing the molar mass of CaO is 56.1 g/mol and knowing that O2 is the limiting reactant, we also know the reaction can produce a maximum of 0,051158645 mol of CaO can be produced. So we can conclude we will produce : 56.1 * 0,051158645 = 2.87g of CaO.