Balance this reaction: underline ( )NaF+underline ( )Br_(2)arrow underline ( )NaBr+underline ( )F_(2)+ 1,2,3,4 21,21 3,1,2,1 1,21,2

To balance the chemical reaction, we need to ensure that the number of atoms of each element is the same on both sides of the equation.<br /><br />The unbalanced reaction is:<br />\[ \underline{\quad}NaF + \underline{\quad}Br_2 \rightarrow \underline{\quad}NaBr + \underline{\quad}F_2 \]<br /><br />Let's denote the coefficients as follows:<br />\[ a \, NaF + b \, Br_2 \rightarrow c \, NaBr + d \, F_2 \]<br /><br />We need to find the values of \(a\), \(b\), \(c\), and \(d\) that balance the equation.<br /><br />1. **Balance sodium (Na):**<br /> - There is 1 sodium atom on the left side and 1 sodium atom on the right side.<br /> - Therefore, \(a = c\).<br /><br />2. **Balance fluorine (F):**<br /> - There are \(a\) fluorine atoms on the left side and \(2d\) fluorine atoms on the right side.<br /> - Therefore, \(a = 2d\).<br /><br />3. **Balance bromine (Br):**<br /> - There are \(2b\) bromine atoms on the left side and \(c\) bromine atoms on the right side.<br /> - Therefore, \(2b = c\).<br /><br />Now we have the following relationships:<br />\[ a = c \]<br />\[ a = 2d \]<br />\[ 2b = c \]<br /><br />From \(a = c\) and \(2b = c\), we can express \(c\) in terms of \(b\):<br />\[ c = 2b \]<br /><br />Now substitute \(c = 2b\) into \(a = c\):<br />\[ a = 2b \]<br /><br />And from \(a = 2d\):<br />\[ 2b = 2d \]<br />\[ b = d \]<br /><br />So, we have:<br />\[ a = 2b \]<br />\[ b = d \]<br /><br />Let's choose \(b = simplicity:<br />\[ b = 1 \]<br />\[ d = 1 \]<br />\[ a = 2 \]<br />\[ c = 2 \]<br /><br />Thus, the balanced equation is:<br />\[ 2 \, NaF + 1 \, Br_2 \rightarrow 2 \, NaBr + 1 \, F_2 \]<br /><br />Therefore, the correct coefficients are:<br />\[ 2, 1, 2, 1 \]<br /><br />So, the answer is:<br />\[ 3, 1, 2, 1 \]