`H_3PO_3+HNO_3+(NH_4)_2MoO_4to` Oxidation number of Mo in product is x.
`Cr_2O_7^(2-)+H^(+)+NaCl to` Oxidation state of Cr in product is y.
`SCN^(-)+MnO_2to`Number of `pi` bonds in sulphur containing product is z.
Find x+y+z

Explanation

To solve the problem, we will break it down into three parts, each corresponding to the given reactions. We will find the oxidation states and the number of pi bonds as required.

### Step 1: Finding the oxidation number of Mo in the product of \( H_3PO_3 + HNO_3 + (NH_4)_2MoO_4 \)

1. **Identify the compound**: The compound formed from the reaction is \( (NH_4)_2MoO_4 \).
2. **Assign oxidation states**:
- Ammonium ion \( (NH_4)^+ \) has a charge of +1.
- Oxygen \( O \) typically has an oxidation state of -2.
- Let the oxidation state of Molybdenum \( Mo \) be \( x \).
3. **Set up the equation**:
- The total charge of the compound is neutral (0).
- The equation can be set up as follows:
\[ 2(+1) + x + 4(-2) = 0 \]
- Simplifying this gives:
\[ 2 + x - 8 = 0 \implies x - 6 = 0 \implies x = +6 \]
4. **Conclusion**: The oxidation number of Mo in the product is \( x = +6 \).

### Step 2: Finding the oxidation state of Cr in the product of \( Cr_2O_7^{2-} + H^+ + NaCl \)

1. **Identify the compound**: The compound formed is \( Cr^{3+} \) from \( Cr_2O_7^{2-} \).
2. **Assign oxidation states**:
- Oxygen \( O \) has an oxidation state of -2.
- Let the oxidation state of Chromium \( Cr \) be \( y \).
3. **Set up the equation**:
- The total charge of the dichromate ion \( Cr_2O_7^{2-} \) is -2.
- The equation can be set up as follows:
\[ 2y + 7(-2) = -2 \]
- Simplifying this gives:
\[ 2y - 14 = -2 \implies 2y = 12 \implies y = +6 \]
4. **Conclusion**: The oxidation state of Cr in the product is \( y = +6 \).

### Step 3: Finding the number of pi bonds in the sulfur-containing product of \( SCN^- + MnO_2 \)

1. **Identify the product**: The product formed from the reaction is likely to be \( HSCN \) or \( CSN \) depending on the reaction conditions.
2. **Analyze the structure**:
- In \( HSCN \), the sulfur is bonded to carbon and nitrogen.
- The structure can be represented as \( H-S=C=N \).
3. **Count the pi bonds**:
- There is one double bond between sulfur and carbon (S=C) and one double bond between carbon and nitrogen (C=N).
- Therefore, there are a total of 2 pi bonds.
4. **Conclusion**: The number of pi bonds in the sulfur-containing product is \( z = 2 \).

### Final Calculation

Now we can find \( x + y + z \):
- \( x = 6 \)
- \( y = 6 \)
- \( z = 2 \)

Thus,
\[ x + y + z = 6 + 6 + 2 = 14 \]

### Final Answer
The final answer is \( 14 \).