Answer to Question #161672 in General Chemistry for Chris Walsh

Question

Practice: Solve using dimensional analysis where appropriate. All numbers must be labeled fully.

1a) Balance: ____ C4H8 (g) + ____ O2 (g) =  ____ CO2 (g) + ____ H2O (g)

Given: If 255 g C4H8 reacts with 325 g O2, then theoretically how many grams of water vapor will form? b) Determine the limiting reactant through calculations.

c) Determine the theoretical mass (g) of water vapor that can be produced.

d) Calculate the mass (g) of the reactant in excess that will be left over when the reaction is complete.

Answer 1

1 a) C₄H₈ + 6O₂ = 4CO₂ + 4H₂O

255g C₄H₈ × $\frac{1 mol C_4H_8}{56.108gC_4H_8}$ = 4.54 mol C₄H₈

325g O₂ × $\frac{1molO_2}{32gO_2}$ = 10.156 mol O₂

b) There are 4.54 mol C₄H₈ and 10.156 mol O₂

To use all the C₄H₈, how many moles of O₂ do we need

4.54 mol C₄H₈ × $\frac{6molO_2}{1 mol C_4H_8}$ = 27.36 mol O₂

To use all the O₂ , how many moles of C₄H₈ do we need

10.156 g O₂ × $\frac{1mole C_4H_8}{6molO_2}$ = 1.692 mol C₄H₈

We have enough C₄H₈ to use all the O₂ but we dont have enough O₂ to use all the C₄H₈ .

O₂ is the limiting reactant .

c) How many moles of H₂O can we use all the O₂

10.156 mol O₂ × $\frac{4molH_2O}{6mol O_2}$ = 6.77 mol H₂O

How many grams of H₂O is this

6.77 × 18.015 = 121.96 g

121.96 g mass of the H₂O we produced

d). C₄H₈ is the Excess reactant

How much are left over

Excess reactant = total reactant - uses reactant

Excess reactant = 4.54 - 1.692

Excess reactant = 2.848 moles of C₄H₈

2.848 moles C₄H₈ × $\frac{56.11g C_4H_8}{1moleC_4H_8}$ = 159.80g

159.80 g C₄H₈ are left over