BR Chapter 1 Stoichiometry Example 1.7

[Ihavesomanyquestions]

---

Members don't see this ad.

BR Chapter 1 Stoichiometry Example 1.7


Based on the following reaction, how many grams of water would form from 0.33 moles C4H10O (g) reacting with an unlimited amount of oxygen gas?


C4H10O(g) + 6O2 (g) ⇌ 4CO2(g) + 5H2O(g)

A. 18.00 grams
B. 24.00 grams
C. 30.00 grams
D. 36.00 grams

Writing out the problem makes it easy but I am confused on their explanation. The ratio of H2O to C4H10O is 5:1 so 1.667 moles of water are formed. Can someone explain this to me? I know that 1.667 comes from dividing 5/3. I do not understand how the ratio is used. Maybe 5:1 then multiply 5 : 1(0.33)?

 

[orangetea]

Is the answer A? if so I can explain to you their explanation if not.. then I must be not understanding it either

 

[sillyjoe]

The ratio of moles of C4H10O to H20 is 1:5

Therefore, with unlimited O2 you would get (.33 moles) * 5 = 1.65 moles H2O

H20 = 18 grams per mole

(18 grams/mole) * 1.65 moles = 29.7 grams = ~30.00 grams

 

[orangetea]

The ratio of moles of C4H10O to H20 is 1:5

Therefore, with unlimited O2 you would get (.33 moles) * 5 = 1.65 moles O2

H20 = 18 grams per mole

(18 grams/mole) * 1.65 moles = 29.7 grams = ~30.00 grams

I need stop with my shady math and not round down when it should be up! thanks 🙂