# Homework 83


Q8.35

Question: The unbalanced equation for the combustion of butane is shown below.

C4H10 + O2 → H2O + CO2

a.) Balance the equation.

b.) Calculate how many moles of C4H10 are required to fully react 17.4 moles of O2.

Solutions:

a.)

1.) Identify the various elements in the equation. Carbon (C), Hydrogen (H), and Oxygen (O) are the species being reacted in this case.

2.) Determine how many atoms of each element are present on the respective sides of the equation.

 Chart 1 C H O O H C # atoms 4 10 2 3 2 1

3.) Chart 1 clearly shows that the equation is not balanced. To balance the equation place integer coefficients infront of the molecules/elements to equal out the equation. Hint: it is usually beneficial to balance oxygen last.

4.) Balanced equation: 2C4H10 + 13O2 → 10H2O + 8CO2

 Chart 2 C H O O H C # atoms 8 20 26 26 20 8

Above is the balanced equation and chart showing the reactants and products balancing out.

b.)

1.) Now using the balanced equation, 2C4H10 + 13O2 → 10H2O + 8CO2 , it is possible to calculated how many moles of C4H10 are required to fully combust 17.4 moles of O2.

2.) The balanced equation tells us that it takes 13 moles of O2 to combust 2 moles of C4H10. Using this information we can create a proportion to solve the problem.

$\frac{2\; mol\; C_{4}H_{10}}{13\; mol\; O_{2}}= \frac{?\; mol\; C_{4}H_{10}}{17.2\; mol\; O_{2}}$

$(2\: mol\; C_{4}H_{10})(17.2\; mol\; O_{2})=(13\; mol\; O_{2})(?\; mol\; C_{4}H_{10})$

$\frac{(2\: mol\; C_{4}H_{10})(17.2\; mol\; O_{2})}{(13\; moles\; O_{2})}=(?\; mol\; C_{4}H_{10})$

$2.65\; mol\; C_{4}H_{10}$

Q9.22

Question: Given the following information calculate the molarity of each solution.

a.) 0.870 mol of H2SO4 in 750. ml of solution

b.) 637.7g of Cu(NO3)2 in 3.83L of solution

c.) 1.51 x 1023 molecules of NaCl in 340 ml of solution

Solutions:

Molarity (M) is defined as the number of moles of a substance per liter of solution

$\frac{moles}{Liter}=Molarity$

a.) $750.\; ml\;*\; (\frac{1\; Liter}{1000\; ml})=\; 0.750L$

$\frac{0.870\; mol\; H_{2}SO_{4}}{.750\; L}=$

1.16M H2SO4

B.)

$(637.7\; g\; Cu(NO_{3})_{2})(\frac{1\; mol\;Cu(NO_{3})_{2} }{187.56\; g})= mol\; Cu(NO_{3})_{2}$

$(\frac{637.7\; g }{187.56\; g})=\;mol\;Cu(NO_{3})_{2}$

$3.400\;mol\;Cu(NO_{3})_{2}$

$\frac{3.400\;mol}{3.83L}=M\;Cu(NO_{3})_{2}$

$.888M\;Cu(NO_{3})_{2}$

C.)

$1.51*10^{23}\; molecules\; of\;NaCl(\frac{mol\; NaCl}{6.022*10^{23}\; molecules})=mol\; NaCl$

$(\frac{1.51*10^{23}\; molecules}{6.022*10^{23}\; molecules})=\; mol\; NaCl$

$0.251\; mol\; NaCl$

$340.\; ml\;*\; (\frac{1\; Liter}{1000\; ml})=\; 0.350L$

$\frac{.251\;mol}{0.350L}=\; M\; NaCl$

$0.717M\; NaCl$

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