Question

balance the equation shown in the video style:
h₂ + cl₂ → __ hcl
○ 1,1,1
○ 2,1,2
○ 1,2,1
○ 2,2,1
question 8
1 pts
balance the equation:
n₂ + h₂ → __ nh₃
○ 1,1,1
○ 1,2,2
○ 1,3,2
○ 2,3,2

Explanation:

Step1: Balance hydrogen and chlorine atoms

For the equation \(H_{2}+Cl_{2}\to HCl\), on the left - hand side, there are 2 hydrogen atoms (\(H\)) and 2 chlorine atoms (\(Cl\)). On the right - hand side, in \(HCl\), there is 1 hydrogen and 1 chlorine atom. If we put a coefficient of 2 in front of \(HCl\), we get \(H_{2}+Cl_{2}\to2HCl\). The number of \(H\) atoms: left - hand side \(= 2\), right - hand side \(=2\times1 = 2\). The number of \(Cl\) atoms: left - hand side \(= 2\), right - hand side \(=2\times1=2\). So the coefficients are \(1,1,2\) (not in the options, but if we consider the general balancing concept).

For the equation \(N_{2}+H_{2}\to NH_{3}\), on the left - hand side, there are 2 nitrogen atoms (\(N\)) and 2 hydrogen atoms (\(H\)). On the right - hand side, in \(NH_{3}\), there is 1 nitrogen and 3 hydrogen atoms.

• Balance nitrogen atoms first. Put a coefficient of 2 in front of \(NH_{3}\), so the equation becomes \(N_{2}+H_{2}\to2NH_{3}\). Now, the number of \(N\) atoms: left - hand side \(=2\), right - hand side \(=2\times1 = 2\).
• Then balance hydrogen atoms. The right - hand side has \(2\times3=6\) hydrogen atoms. So put a coefficient of 3 in front of \(H_{2}\) on the left - hand side. The equation is \(N_{2}+3H_{2}\to2NH_{3}\). The number of \(H\) atoms: left - hand side \(=3\times2 = 6\), right - hand side \(=2\times3=6\).

Answer:

For \(H_{2}+Cl_{2}\to HCl\), the balanced equation is \(1H_{2}+1Cl_{2}\to2HCl\) (but if we assume a mis - print in options and follow the balancing rules). For \(N_{2}+H_{2}\to NH_{3}\), the balanced equation is \(1N_{2}+3H_{2}\to2NH_{3}\), so the answer is \(1,3,2\) (i.e., the option \(1,3,2\) for the \(N_{2}+H_{2}\to NH_{3}\) equation).