Question

hester, treyvon
dtal - ch - term 1 - unit 5a assessment 25 - 26
8 of 15
ca(no₃)₂+h₃po₄→ca₃(po₄)₂+hno₃
when this equation is balanced, the coefficient in front of h₃po₄ is -
1
2
3
4

Explanation:

Step1: Balance Ca atoms

On the right - hand side, there are 3 Ca atoms in \(Ca_3(PO_4)_2\). So, we put a coefficient of 3 in front of \(Ca(NO_3)_2\) on the left - hand side. The equation becomes \(3Ca(NO_3)_2+\text{?}H_3PO_4
ightarrow Ca_3(PO_4)_2 +\text{?}HNO_3\)

Step2: Balance \(NO_3\) groups

Since there are 6 \(NO_3\) groups in \(3Ca(NO_3)_2\), we put a coefficient of 6 in front of \(HNO_3\) on the right - hand side. The equation is now \(3Ca(NO_3)_2+\text{?}H_3PO_4
ightarrow Ca_3(PO_4)_2+6HNO_3\)

Step3: Balance \(PO_4\) groups and \(H\) atoms

On the right - hand side, there are 2 \(PO_4\) groups in \(Ca_3(PO_4)_2\). So, we put a coefficient of 2 in front of \(H_3PO_4\) on the left - hand side.
For \(H\) atoms: In \(2H_3PO_4\), there are \(2\times3 = 6\) \(H\) atoms. In \(6HNO_3\), there are \(6\times1=6\) \(H\) atoms.
The balanced equation is \(3Ca(NO_3)_2 + 2H_3PO_4
ightarrow Ca_3(PO_4)_2+6HNO_3\)

Answer:

B. 2