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# Balancing of Chemical Reactions using the Algebraic Method

Balance the chemical reactions using the Algebraic Method
Check your solution using the Inspection Method

(1.) ${SO_2}_{(g)} + {O_2}_{(g)} \rightarrow {SO_3}_{(g)}$

$xSO_2 + yO_2 \rightarrow SO_3 \\[3ex] S = x ~~~~~~~~~~~~\:\:\:\:\:\:\: S = 1 \\[3ex] O = 2x + 2y \:\:\:\:\:\:\: O = 3 \\[3ex] x = 1 \\[3ex] 2x + 2y = 3 \\[3ex] 2(1) + 2y = 3 \\[3ex] 2 + 2y = 3 \\[3ex] 2y = 3 - 2 \\[3ex] 2y = 1 \\[3ex] y = \dfrac{1}{2} \\[5ex] Substitute\:\:the\:\:values\:\:back\:\:in\:\:the\:\:equation \\[3ex] 1SO_2 + \dfrac{1}{2}O_2 \rightarrow SO_3 \\[5ex] LCD = 2 \\[3ex] Multiply\:\:each\:\:reactant\:\:and\:\:each\:\:product\:\:by\:\:the\:\:LCD \\[3ex] 2 * 1 * SO_2 + 2 * \dfrac{1}{2} * O_2 \rightarrow 2 * SO_3 \\[5ex] \therefore 2SO_2 + O_2 \rightarrow 2SO_3 \\[3ex]$ Check
 $\underline{Reactants} \\[3ex] S = 2 \\[3ex] O = 2(2) + 2 = 4 + 2 = 6$ $\underline{Products} \\[3ex] S = 2 \\[3ex] O = 2 * 3 = 6$
(2.) ${C_{4}H_{10}}_{(g)} + {O_2}_{(g)} \rightarrow {CO_2}_{(g)} + {H_{2}O}_{(g)}$

$xC_{4}H_{10} + yO_2 \rightarrow pCO_2 + H_{2}O \\[3ex] C = 4x ~~~\:\:\:\:\:\:\: C = p \\[3ex] H = 10x \:\:\:\:\:\:\: H = 2 \\[3ex] O = 2y ~~~\:\:\:\:\:\:\: O = 2p + 1 \\[3ex] 4x = p ...eqn.(1) \\[3ex] 10x = 2 ...eqn.(2) \\[3ex] 2y = 2p + 1 ...eqn.(3) \\[3ex] From\:\:eqn.(2):\:\: x = \dfrac{2}{10} = \dfrac{1}{5} \\[5ex] From\:\:eqn.(1):\:\: 4 * \dfrac{1}{5} = p \\[5ex] p = \dfrac{4}{5} \\[5ex] From\:\:eqn.(3):\:\: 2y = 2 * \dfrac{4}{5} + 1 \\[5ex] 2y = \dfrac{8}{5} + \dfrac{5}{5} \\[5ex] 2y = \dfrac{8 + 5}{5} = \dfrac{13}{5} \\[5ex] y = \dfrac{1}{2} * \dfrac{13}{5} \\[5ex] y = \dfrac{13}{10} \\[5ex] Substitute\:\:the\:\:values\:\:back\:\:in\:\:the\:\:equation \\[3ex] \dfrac{1}{5}C_{4}H_{10} + \dfrac{13}{10}O_2 \rightarrow \dfrac{4}{5}CO_2 + H_{2}O \\[5ex] LCD = 10 \\[3ex] Multiply\:\:each\:\:reactant\:\:and\:\:each\:\:product\:\:by\:\:the\:\:LCD \\[3ex] 10 * \dfrac{1}{5}C_{4}H_{10} + 10 * \dfrac{13}{10}O_2 \rightarrow 10 * \dfrac{4}{5}CO_2 + 10 * 1H_{2}O \\[5ex] \therefore 2C_{4}H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_{2}O \\[3ex]$ Check
 $\underline{Reactants} \\[3ex] C = 2 * 4 = 8 \\[3ex] H = 2 * 10 = 20 \\[3ex] O = 13 * 2 = 26$ $\underline{Products} \\[3ex] C = 8 \\[3ex] H = 10 * 2 = 20 \\[3ex] O = 8 * 2 + 10 = 16 + 10 = 26$
(3.) $Cu_{(s)} + {HNO_3}_{(aq)} \rightarrow {Cu(NO_3)_2}_{(aq)} + NO_{(g)} + {H_{2}O}_{(l)}$

$xCu + yHNO_3 \rightarrow pCu(NO_3)_2 + dNO + H_{2}O \\[3ex] Cu = x ~~~\:\:\:\:\:\:\: Cu = p \\[3ex] H = y ~~~~~\:\:\:\:\:\:\: H = 2 \\[3ex] N = y ~~~~~\:\:\:\:\:\:\: N = 2p + d \\[3ex] O = 3y ~~~\:\:\:\:\:\:\: O = 6p + d + 1 \\[3ex] x = p ...eqn.(1) \\[3ex] y = 2 ...eqn.(2) \\[3ex] y = 2p + d ...eqn.(3) \\[3ex] 3y = 6p + d + 1 ...eqn.(4) \\[3ex] Substitute\:\:the\:\:value\:\:of\:\:y\:\:in\:\:eqn.(3)\:\:and\:\:eqn.(4) \\[3ex] From\:\:eqn.(3):\:\: 2 = 2p + d \\[3ex] 2p + d = 2 ...eqn.(3) \\[3ex] From\:\:eqn.(4):\:\: 3(2) = 6p + d + 1 \\[3ex] 6 = 6p + d + 1 \\[3ex] 6p + d + 1 = 6 ...eqn.(4) \\[3ex] eqn.(4) - eqn.(3) \implies (6p + d + 1) - (2p + d) = 6 - 2 \\[3ex] 6p + d + 1 - 2p - d = 4 \\[3ex] 4p = 4 - 1 \\[3ex] 4p = 3 \\[3ex] p = \dfrac{3}{4} \\[5ex] Substitute\:\:the\:\:value\:\:of\:\:p\:\:in\:\:eqn.(1)\:\:and\:\:eqn.(3) \\[3ex] From\:\:eqn.(1):\:\: x = \dfrac{3}{4} \\[5ex] From\:\:eqn.(3):\:\: d = 2 - 2p \\[3ex] d = 2 - 2\left(\dfrac{3}{4}\right) \\[5ex] d = 2 - \dfrac{3}{2} \\[5ex] d = \dfrac{4}{2} - \dfrac{3}{2} \\[5ex] d = \dfrac{1}{2} \\[5ex] Substitute\:\:the\:\:values\:\:back\:\:in\:\:the\:\:equation \\[3ex] \dfrac{3}{4}Cu + 2HNO_3 \rightarrow \dfrac{3}{4}Cu(NO_3)_2 + \dfrac{1}{2}NO + H_{2}O \\[5ex] LCD = 4 \\[3ex] Multiply\:\:each\:\:reactant\:\:and\:\:each\:\:product\:\:by\:\:the\:\:LCD \\[3ex] 4 * \dfrac{3}{4}Cu + 4 * 2HNO_3 \rightarrow 4 * \dfrac{3}{4}Cu(NO_3)_2 + 4 * \dfrac{1}{2}NO + 4 * 1H_{2}O \\[5ex] \therefore 3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 2NO + 4H_{2}O \\[3ex]$ Check
 $\underline{Reactants} \\[3ex] Cu = 3 \\[3ex] H = 8 \\[3ex] N = 8 \\[3ex] O = 8 * 3 = 24$ $\underline{Products} \\[3ex] Cu = 3 \\[3ex] H = 4 * 2 = 8 \\[3ex] N = 3 * 2 + 2 = 6 + 2 = 8 \\[3ex] O = 3 * 3 * 2 + 2 + 4 = 18 + 2 + 4 = 24$
(4.) ${Fe(OH)_3}_{(s)} + {H_2SO_4}_{(aq)} \rightarrow {Fe_{2}(SO_4)_3}_{(s)} + {H_{2}O}_{(l)}$

$xFe(OH)_3 + yH_{2}SO_4 \rightarrow pFe_{2}(SO_4)_3 + H_{2}O \\[3ex] Fe = x ~~~~~~~~~~~~~~~\:\:\:\:\:\:\: Fe = 2p \\[3ex] O = 3x + 4y ~~~~~~\:\:\:\:\:\:\: O = 12p + 1 \\[3ex] H = 3x + 2y ~~~~~\:\:\:\:\:\:\: H = 2 \\[3ex] S = y ~~~~~~~~~~~~~~~~~\:\:\:\:\:\:\: S = 3p \\[3ex] x = 2p ...eqn.(1) \\[3ex] 3x + 4y = 12p + 1 ...eqn.(2) \\[3ex] 3x + 2y = 2 ...eqn.(3) \\[3ex] y = 3p ...eqn.(4) \\[3ex] Substitute\:\:eqn.(1)\:\:and\:\:eqn.(4)\:\:in\:\:eqn.(2) \\[3ex] Substitute\:\:2p\:\:for\:\:x\:\:and\:\:3p\:\:for\:\:y\:\:in\:\:eqn.(2) \\[3ex] 3(2p) + 4(3p) = 12p + 1 \\[3ex] 6p + 12p = 12p + 1 \\[3ex] 18p - 12p = 1 \\[3ex] 6p = 1 \\[3ex] p = \dfrac{1}{6} \\[5ex] From\:\:eqn.(1):\:\: x = 2\left(\dfrac{1}{6}\right) = \dfrac{1}{3} \\[5ex] From\:\:eqn.(4):\:\: y = 3\left(\dfrac{1}{6}\right) = \dfrac{1}{2} \\[5ex] Substitute\:\:the\:\:values\:\:back\:\:in\:\:the\:\:equation \\[3ex] \dfrac{1}{3}Fe(OH)_3 + \dfrac{1}{2}H_{2}SO_4 \rightarrow \dfrac{1}{6}Fe_{2}(SO_4)_3 + H_{2}O \\[5ex] LCD = 6 \\[3ex] Multiply\:\:each\:\:reactant\:\:and\:\:each\:\:product\:\:by\:\:the\:\:LCD \\[3ex] 6 * \dfrac{1}{3}Fe(OH)_3 + 6 * \dfrac{1}{2}H_{2}SO_4 \rightarrow 6 * \dfrac{1}{6}Fe_{2}(SO_4)_3 + 6 * 1H_{2}O \\[5ex] \therefore 2Fe(OH)_3 + 3H_{2}SO_4 \rightarrow Fe_{2}(SO_4)_3 + 6H_{2}O \\[3ex]$ Check
 $\underline{Reactants} \\[3ex] Fe = 2 \\[3ex] O = 2 * 3 + 3 * 4 = 6 + 12 = 18 \\[3ex] H = 2 * 3 + 3 * 2 = 6 + 6 = 12 \\[3ex] S = 3$ $\underline{Products} \\[3ex] Fe = 2 \\[3ex] O = 4 * 3 + 6 = 12 + 6 = 18 \\[3ex] H = 6 * 2 = 12 \\[3ex] S = 3$