45 - (CHEMI/10.HL. Winter 2022 (Q3) - Stoichiometric Relationship How many moles of carbon dioxide are produced by the complete combustion of 7.0 g of ethene, C_(2)H_(4)(g) M_(r)=28 A. 0.25 B. 0.5 C. 0.75 D. 1.0

To find the number of moles of carbon dioxide produced by the complete combustion of 7.0 g of ethene, we need to use the stoichiometric relationship between the reactants and products in the balanced chemical equation for the combustion of ethene.<br /><br />The balanced chemical equation for the complete combustion of ethene is:<br /><br />$C_{2}H_{4}(g) + 3O_{2}(g) \rightarrow 2CO_{2}(g) + 2H_{2}O(l)$<br /><br />From the balanced equation, we can see that 1 mole of ethene reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide.<br /><br />Given that the molar mass of ethene ($C_{2}H_{4}$) is 28 g/mol, we can calculate the number of moles of ethene in 7.0 g:<br /><br />Number of moles of ethene = $\frac{7.0 \text{ g}}{28 \text{ g/mol}} = 0.25 \text{ moles}$<br /><br />According to the balanced equation, 1 mole of ethene produces 2 moles of carbon dioxide. Therefore, 0.25 moles of ethene will produce:<br /><br />Number of moles of carbon dioxide = $0.25 \text{ moles} \times 2 = 0.5 \text{ moles}$<br /><br />Therefore, the correct answer is B. 0.5 moles of carbon dioxide are produced by the complete combustion of 7.0 g of ethene.