Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

practice problems! now you try balancing some chemical equations for yo…

Question

practice problems!
now you try balancing some chemical equations for yourself!

  1. __ k + h₂o → koh + __ h₂
  2. __ al + hcl → alcl₃ + __ h
  3. __ agno₃ + ni → ni(no₃)₂ + __ ag
  4. __ h₃po₄ + koh → k₃po₄ + __ h₂o

Explanation:

Step1: Balance hydrogen and oxygen in the first equation

For the equation \(K + H_{2}O
ightarrow KOH + H_{2}\), we first look at the hydrogen atoms. There are 2 hydrogen atoms in \(H_{2}O\) and 3 hydrogen atoms on the right - hand side (\(1\) in \(KOH\) and \(2\) in \(H_{2}\)). If we put a coefficient of \(2\) in front of \(H_{2}O\), we have \(2H_{2}O\) (4 hydrogen atoms). Then, if we put a coefficient of \(2\) in front of \(KOH\) (2 hydrogen atoms from \(KOH\)) and a coefficient of \(1\) in front of \(H_{2}\), the hydrogen atoms are balanced. Now for potassium atoms, if we put a coefficient of \(2\) in front of \(K\), the equation \(2K + 2H_{2}O=2KOH + H_{2}\) is balanced.

Step2: Balance chlorine and hydrogen in the second equation

For the equation \(Al+HCl
ightarrow AlCl_{3}+H_{2}\), we first balance the chlorine atoms. Since there are 3 chlorine atoms in \(AlCl_{3}\), we put a coefficient of \(3\) in front of \(HCl\). Now for hydrogen atoms, we have 3 hydrogen atoms from \(HCl\). To balance the hydrogen atoms (since we want \(H_{2}\)), we multiply \(HCl\) by \(6\) (so we have 6 hydrogen atoms). Then, to balance aluminium, we put a coefficient of \(2\) in front of \(Al\). The balanced equation is \(2Al + 6HCl=2AlCl_{3}+3H_{2}\)

Step3: Balance nitrate and metal atoms in the third equation

For the equation \(AgNO_{3}+Ni
ightarrow Ni(NO_{3})_{2}+Ag\), we first balance the nitrate ions. Since there are 2 nitrate ions in \(Ni(NO_{3})_{2}\), we put a coefficient of \(2\) in front of \(AgNO_{3}\). Then, to balance the silver atoms, we put a coefficient of \(2\) in front of \(Ag\). The balanced equation is \(2AgNO_{3}+Ni = Ni(NO_{3})_{2}+2Ag\)

Step4: Balance potassium and hydrogen in the fourth equation

For the equation \(H_{3}PO_{4}+KOH
ightarrow K_{3}PO_{4}+H_{2}O\), we first balance the potassium atoms. Since there are 3 potassium atoms in \(K_{3}PO_{4}\), we put a coefficient of \(3\) in front of \(KOH\). Now for hydrogen atoms, we have \(3\) hydrogen atoms from \(H_{3}PO_{4}\) and \(3\) hydrogen atoms from \(KOH\) (total \(6\) hydrogen atoms on the left). On the right - hand side, in \(H_{2}O\), if we put a coefficient of \(3\) in front of \(H_{2}O\) (6 hydrogen atoms), the equation \(H_{3}PO_{4}+3KOH = K_{3}PO_{4}+3H_{2}O\) is balanced

Answer:

  1. \(2K + 2H_{2}O=2KOH + H_{2}\)
  2. \(2Al + 6HCl=2AlCl_{3}+3H_{2}\)
  3. \(2AgNO_{3}+Ni = Ni(NO_{3})_{2}+2Ag\)
  4. \(H_{3}PO_{4}+3KOH = K_{3}PO_{4}+3H_{2}O\)