Homework 83
- Page ID
- 29266
Q8.35
The unbalanced equation for the combustion of butane is shown below.
\[C_4H_{10} + O_2 \rightarrow H_2O + CO_2\]
- Balance the equation.
- Calculate how many moles of \(C_4H_{10}\) are required to fully react 17.4 moles of \(O_2\).
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 | Reactants | Products | |||||
Elements | C | H | O | O | H | C | |
# atoms | 4 | 10 | 2 | 3 | 2 | 1 |
3.) Chart 1 shows that the equation is not balanced. To balance the equation place integer coefficients infront of the molecules/elements to equal out the equation. It is not uncommon to have to try several different combination of coefficients to balance the equation Hint: it is usually beneficial to balance oxygen last.
4.) Balanced equation: 2C_{4}H_{10} + 13O_{2} → 10H_{2}O + 8CO_{2}
Chart 2 | Reactants | Products | |||||
Elements | 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, 2C_{4}H_{10} + 13O_{2} → 10H_{2}O + 8CO_{2 , }it is possible to calculated how many moles of C_{4}H_{10 }are required to fully combust 17.4 moles of O_{2}.
2.) The balanced equation tells us that it takes 13.0 moles of O_{2 }to combust 2.00 moles of C_{4}H_{10. }Using this information we can create a proportion to solve the problem.
\[\dfrac{2.00\; mol\; C_{4}H_{10}}{13.0\; mol\; O_{2}}= \dfrac{?\; mol\; C_{4}H_{10}}{17.2\; mol\; O_{2}}\]
\[(2.00\: mol\; C_{4}H_{10})(17.2\; mol\; O_{2})=(13\; mol\; O_{2})(?\; mol\; C_{4}H_{10})\]
\[\dfrac{(2.00\: mol\; C_{4}H_{10})(17.2\; mol\; O_{2})}{(13.0\; moles\; O_{2})}=(?\; mol\; C_{4}H_{10})\]
\[2.65\; mol\; C_{4}H_{10}\]
Q9.22
Given the following information calculate the molarity of each solution.
- \(0.870\; mol\) of \(H_2SO_4\) in 750.0 ml of solution
- \(637.7\; g\) of \(Cu(NO_3)_2\) in 3.83 L of solution
- \(1.51 \times 10^{23}\) 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.
\[\dfrac{moles}{Liter}=Molarity\]
a.) \(0.870\; mol\) of \(H_2SO_4\) in \(750.0 \;ml\) of solution
\[750.0\; \cancel{ml}\;*\; \left(\dfrac{1\; L}{1000\; \cancel{ml}}\right)=\; 0.750\;L\]
\[\dfrac{0.870\; mol\; H_{2}SO_{4}}{0.750\; L}=M\; H_{2}SO_{4}\]
\[1.16M\; H_{2}SO_{4}\]
B.) \(637.7\;g\) of \(Cu(NO_3)_2\) in 3.83 L of solution
\[(637.7\; g\; Cu(NO_{3})_{2})\left(\dfrac{1\; mol\;Cu(NO_{3})_{2} }{187.56\; g}\right)= mol\; Cu(NO_{3})_{2}\]
\[(\dfrac{637.7\; g }{187.56\; g/mol})=3.400\;mol\;Cu(NO_{3})_{2}\]
\[\dfrac{3.400\;mol}{3.83L}=M\;Cu(NO_{3})_{2}\]
\[0.888M\;Cu(NO_{3})_{2}\]
C.) 1.51 x 10^{23 }molecules of NaCl in 340 ml of solution
\[1.51 \times 10^{23}\; molecules\; of\;NaCl(\dfrac{mol\; NaCl}{6.022 \times 10^{23}\; molecules})=mol\; NaCl\]
\[\left(\dfrac{1.51 \times 10^{23}\; molecules}{6.022*10^{23}\; molecules}\right)=\; mol\; NaCl\]
\[0.251\; mol\; NaCl\]
\[340.\; ml\;*\; (\dfrac{1\; Liter}{1000\; ml})=\; 0.350L\]
\[\dfrac{0.251\;mol}{0.350L}=\; M\; NaCl\]
\[0.717M\; NaCl\]