× | C late the mass of water How many moles of hydro + tlist/5601995 Search ome

Question

× | C late the mass of water How many moles of hydro + tlist/5601995 Search ome work- Mozilla Firefox nView?offset- nextassignmentProblemID- 87819523 previous | 19 of 23 I next 1. Convert from grams of compound X to moles of compound X using the molar mass of compound X 2. Convert from moles of compound X to moles of compound Y using the coeffiients in the balanced chemical equation. 3. Convert from moles of compound Y to grams of compound Y using the molar mass of compound Y Part B Calculate the mass of water produced when 2.40 g of butane reacts with excess oxygen. Express your answer to three significant figures and include the appropriate units. Hints Submit My Answers Give Up Incorrect; Try Again; 5 attempts remaining Part C Calculate the mass of butane needed to produce 59 8 g of carbon dioxide Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

B)


Molar mass of C4H10,
MM = 4*MM(C) + 10*MM(H)
= 4*12.01 + 10*1.008
= 58.12 g/mol


mass(C4H10)= 2.4 g

number of mol of C4H10,
n = mass of C4H10/molar mass of C4H10
=(2.4 g)/(58.12 g/mol)
= 4.129*10^-2 mol
Balanced chemical equation is:
2 C4H10 + 13 O2 ---> 8 H2O + 10 CO2


Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol

According to balanced equation
mol of H2O formed = (8/2)* moles of C4H10
= (8/2)*0.0413
= 0.1652 mol


mass of H2O = number of mol * molar mass
= 0.1652*18.02
= 2.98 g
Answer: 2.98 g

C)

Molar mass of CO2,
MM = 1*MM(C) + 2*MM(O)
= 1*12.01 + 2*16.0
= 44.01 g/mol

mass of CO2 = 59.8 g
molar mass of CO2 = 44.01 g/mol
mol of CO2 = (mass)/(molar mass)
= 59.8/44.01
= 1.3588 mol


According to balanced equation
mol of C4H1O required = (2/10)* moles of CO2
= (2/10)*1.3588
= 0.2718 mol



mass of C4H1O = number of mol * molar mass
= 0.2718*65.048
= 17.7 g

Answer: 17.7 g

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.