Q1) (20pts) Boron sulfide reacts violently with water to form boric acid and hydrogen sulfide gas (the smell of rotten eggs). The unbalanced equation is B_(2)S_(3)+H_(2)Oarrow H_(3)BO_(3)+H_(2)S a) Write a vector equation and matrix equation of the linear system corresponding to the chemical equation given above. b) Balance the chemical equation given above using the linear system solution approach (that we learned in CH-1) Show your work clearly.

a) To write a vector equation and matrix equation for the linear system corresponding to the chemical equation, we need to represent the number of moles of each substance involved in the reaction.<br /><br />Let's define the following variables:<br />x = number of moles of $B_{2}S_{3}$<br />y = number of moles of $H_{2}O$<br />z = number of moles of $H_{3}BO_{3}$<br />w = number of moles of $H_{2}S$<br /><br />The vector equation representing the chemical equation is:<br />$\begin{bmatrix} x \\ y \\ z \\ w \end{bmatrix} = \begin{bmatrix} 1 \\ a \\ b \\ c \end{bmatrix} \begin{bmatrix} 2 \\ 3 \\ d \\ e \end{bmatrix} \begin{bmatrix} 0 \\ f \\ g \end{bmatrix} \begin{bmatrix} 0 \\ h \end{bmatrix}$<br /><br />The matrix equation representing the chemical equation is:<br />$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ w \end{bmatrix} = \begin{bmatrix} 1 \\ a \\ b \\ c \end{bmatrix} \begin{bmatrix} 2 \\ 3 \\ d \\ e \end{bmatrix} \begin{bmatrix} 0 \\ f \\ g \end{bmatrix} \begin{bmatrix} 0 \\ h \end{bmatrix}$<br /><br />b) To balance the chemical equation using the linear system solution approach, we need to find the values of a, b, c, d, e, f, g, and h that satisfy the equation.<br /><br />From the unbalanced equation, we can see that the number of boron atoms on both sides should be equal. Therefore, we have:<br />2x = z<br /><br />The number of sulfur atoms on both sides should also be equal. Therefore, we have:<br />3x = w<br /><br />The number of hydrogen atoms on both sides should be equal. Therefore, we have:<br />2y = 3z + 2w<br /><br />The number of oxygen atoms on both sides should be equal. Therefore, we have:<br />3y = 3b<br /><br />Now, we can solve this system of equations to find the values of x, y, z, and w.<br /><br />From the first equation, we have:<br />z = 2x<br /><br />From the second equation, we have:<br />w = 3x<br /><br />Substituting these values into the third equation, we get:<br />2y = 3(2x) + 2(3x)<br />2y = 12x + 6x<br />2y = 18x<br /><br />Dividing both sides by 2, we have:<br />y = 9x<br /><br />Substituting this value into the fourth equation, we get:<br />3(9x) = 3b<br />27x = 3b<br /><br />Dividing both sides by 3, we have:<br />b = 9x<br /><br />Now, we can substitute the values of z and b into the balanced equation:<br />$B_{2}S_{3}+9H_{2}O\rightarrow 2H_{3}BO_{3}+3H_{2}S$<br /><br />Therefore, the balanced chemical equation is:<br />$B_{2}S_{3}+9H_{2}O\rightarrow 2H_{3}BO_{3}+3H_{2}S$