Question

O STOICHIOMETRY Limiting reactants Kaila Gaseous butane CH, CH2), CH3 will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2 and gaseous water (H20). Suppose 17. 9 of butane is mixed with 31.2 g of oxygen. Calculate the minimum mass of butane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits. xs ? Explanation) Check

Answer 1

Solutions First we have to balanced and complete reaction. CH₃ (CH2 CH3) + O2 H₂O + CO2 Ca Hjo +02 H2O + CO2 First balance the Hydrogen (H) in both side CAHJO + 02 - 5 H₂O + 602 Then balance the carbon (C) in both side C4H10 +02 5H20 + 4602 Then Balance the On in both side C4H10 + 1302 - 5H20 + 4CO2 2 CAHIO + 1302 - 10 H₂O + CO2 This is balanced equation Molecular mass of Caltio= (1244) + (10x15 g (mol - 50.0 g 1 moe Cahyo = 58.0 g/mol

Molecular mass of o2 = 2 h = 32 g/ maliy Here a male of Canso (Butance) react with 13 more of o2 amoll of cato = 250 g hot = 116g 13 male of O2 = 1342 = 416g Here, 4169 O2 react with = 1169 of CalHIO So, ag On react with = 116 g of CAHIO giver = 31.29 Os react with Cel6.x312) да оf сано = 8.7g of Canso Ciren mass of Calo = 17.0g, So, leicht over mass of Calso = (17.0 8.7) g So, le d o sta de 8.39 rolat. Minimum mass of Butane left ove reaction = 8.39 Ans: ARS 8.39