Question

C_{2}H_{6}(g) + O_{2}(g) -->
CO_{2}(g) + H_{2}O(g)

(a) Consider the unbalanced equation above. How many grams of
O_{2} are required to react with 15.0 g of
C_{2}H_{6}? Use at least as many significant
figures in your molar masses as in the data given

g O_{2}

(b) What mass of CO_{2} is produced?

g CO_{2}

(c) What mass of H_{2}O is produced?

g H_{2}O

Answer #1

**Molar mass of C2H6,**

**MM = 2*MM(C) + 6*MM(H)**

**= 2*12.01 + 6*1.008**

**= 30.068 g/mol**

**mass(C2H6)= 15.0 g**

**number of mol of C2H6,**

**n = mass of C2H6/molar mass of C2H6**

**=(15.0 g)/(30.068 g/mol)**

**= 0.4989 mol**

**Balanced chemical equation is:**

**2 C2H6 + 7 O2 ---> 4 CO2 + 6 H2O**

**a)**

**mol of O2 required = (7/2)*moles of C2H6**

**= (7/2)*).4989 mol**

**= 1.746 mol**

**molar mass of O2 = 32 g/mol**

**mass of O2 = number of mol * molar mass**

**= 1.746 mol * 32 g/mol**

**= 55.9 g**

**Answer: 55.9 g**

**b)**

**Molar mass of CO2,**

**MM = 1*MM(C) + 2*MM(O)**

**= 1*12.01 + 2*16.0**

**= 44.01 g/mol**

**According to balanced equation**

**mol of CO2 formed = (4/2)* moles of C2H6**

**= (4/2)*0.498869**

**= 0.997738 mol**

**mass of CO2 = number of mol * molar mass**

**= 0.9977*44.01**

**= 43.9 g**

**Answer: 43.9 g**

**c)**

**According to balanced equation**

**mol of H2O formed = (4/2)* moles of C2H6**

**= (4/2)*0.498869**

**= 0.997738 mol**

**mass of H2O = number of mol * molar mass**

**= 0.9977*18.02**

**= 18.0 g**

**Answer: 18.0 g**

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Part B:
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