Question

grams A to moles A to moles B to grams B 2 C4H10 (l) + 13 O2(g) --> 8 CO2(g) + 10 H2O(g) How many moles of each product will form if 8.12 g of butane (C4H10) reacts completely to form products? Assume there is more than enough of the other reactant. How many grams of CO2 will form?

Answer #1

**1)**

**Molar mass of C4H10,**

**MM = 4*MM(C) + 10*MM(H)**

**= 4*12.01 + 10*1.008**

**= 58.12 g/mol**

**mass(C4H10)= 8.12 g**

**number of mol of C4H10,**

**n = mass of C4H10/molar mass of C4H10**

**=(8.12 g)/(58.12 g/mol)**

**= 0.1397 mol**

**Balanced chemical equation is:**

**2 C4H10 + 13 O2 ---> 8 CO2 + 10 H2O**

**According to balanced equation**

**mol of CO2 formed = (8/2)* moles of C4H10**

**= (8/2)*0.1397**

**= 0.5588 mol**

**Answer: 0.559 moles of CO2 formed**

**2)**

**According to balanced equation**

**mol of H2O formed = (10/2)* moles of C4H10**

**= (10/2)*0.1397**

**= 0.6986 mol**

**Answer: 0.699 moles of H2O formed**

**3)**

**Molar mass of CO2,**

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

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

**= 44.01 g/mol**

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

**= 0.5588*44.01**

**= 24.6 g**

**Answer: 24.6 g of CO2 formed**

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