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(B) To balance the equation $x KNO_3 + y C_{12}H_{22}O_{11} \longrightarrow p N_2 + q CO_2 + r H_2O + s K_2CO_3$:$1$. Balance $K$: $x = 2s$$2$. Balanc

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$x=48, y=5, p=24, q=36, r=55, s=24$

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(B) To balance the equation $x KNO_3 + y C_{12}H_{22}O_{11} \longrightarrow p N_2 + q CO_2 + r H_2O + s K_2CO_3$:$1$. Balance $K$: $x = 2s$$2$. Balance $N$: $x = 2p$$3$. Balance $C$: $12y = q + s$$4$. Balance $H$: $22y = 2r \implies 11y = r$$5$. Balance $O$: $3x + 11y = 2q + r + 3s$Substituting the values from option $(b)$: $x=48, y=5, p=24, q=36, r=55, s=24$.$48 KNO_3 + 5 C_{12}H_{22}O_{11} \longrightarrow 24 N_2 + 36 CO_2 + 55 H_2O + 24 K_2CO_3$Checking atoms:$K: 48 = 2(24) = 48$ (Balanced)$N: 48 = 2(24) = 48$ (Balanced)$C: 12(5) = 60; 36 + 24 = 60$ (Balanced)$H: 22(5) = 110; 55(2) = 110$ (Balanced)$O: 3(48) + 11(5) = 144 + 55 = 199; 2(36) + 55 + 3(24) = 72 + 55 + 72 = 199$ (Balanced)Thus, option $(b)$ is correct.

Balance the following chemical equation by choosing the correct option for the coefficients $x, y, p, q, r,$ and $s$.
$x KNO_3 + y C_{12}H_{22}O_{11} \longrightarrow p N_2 + q CO_2 + r H_2O + s K_2CO_3$

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Balance the following chemical equation by choosing the correct option for the coefficients $x, y, p, q, r,$ and $s$.$x KNO_3 + y C_{12}H_{22}O_{11} \longrightarrow p N_2 + q CO_2 + r H_2O + s K_2CO_3$