Uranium is isolated from its ore by dissolving it as `UO_(2)(NO_(3))_(2)` and separating it as solid `UO_(2)(C_(2)O_(4)).xH_(2)O " "A 1.0 g` sample of ore on treatment with nitric acid yielded `1.48 g UO_(2)(NO_(3))_(2)` which on further treatment with `0.4 g Na_(2)C_(2)O_(4)` yielded `1.23 g " "UO_(2)" "(C_(2)O_(4)).xH_(2)O`. Determine weight percentage of uranium in the original sample and `x` :

Explanation

To solve the problem step by step, we need to determine the weight percentage of uranium in the original sample and the value of \( x \) in the compound \( UO_2(C_2O_4) \cdot xH_2O \).

### Step 1: Calculate the mass of uranium in \( UO_2(NO_3)_2 \)

We know that a 1.48 g sample of \( UO_2(NO_3)_2 \) was obtained from the ore. The molar mass of \( UO_2(NO_3)_2 \) can be calculated as follows:

- Molar mass of Uranium (U) = 238 g/mol
- Molar mass of Oxygen (O) = 16 g/mol
- Molar mass of Nitrogen (N) = 14 g/mol

The molar mass of \( UO_2(NO_3)_2 \):
\[ \text{Molar mass} = 238 + 2(16) + 2(14 + 3(16)) = 238 + 32 + 2(14 + 48) = 238 + 32 + 2(62) = 238 + 32 + 124 = 394 \text{ g/mol} \]

Now, we can find the mass of uranium in the 1.48 g of \( UO_2(NO_3)_2 \):
\[ \text{Mass of U} = \left( \frac{238 \text{ g/mol}}{394 \text{ g/mol}} \right) \times 1.48 \text{ g} = 0.894 \text{ g} \]

### Step 2: Calculate the weight percentage of uranium in the original sample

The original sample of ore weighed 1.0 g. The weight percentage of uranium in the ore can be calculated as follows:
\[ \text{Weight percentage of U} = \left( \frac{\text{Mass of U}}{\text{Mass of ore}} \right) \times 100 = \left( \frac{0.894 \text{ g}}{1.0 \text{ g}} \right) \times 100 = 89.4\% \]

### Step 3: Determine the number of moles of \( Na_2C_2O_4 \)

Next, we need to find the number of moles of \( Na_2C_2O_4 \) used in the reaction. Given that 0.4 g of \( Na_2C_2O_4 \) was used, we calculate its molar mass:
- Molar mass of Sodium (Na) = 23 g/mol
- Molar mass of Carbon (C) = 12 g/mol
- Molar mass of Oxygen (O) = 16 g/mol

The molar mass of \( Na_2C_2O_4 \):
\[ \text{Molar mass} = 2(23) + 2(12) + 4(16) = 46 + 24 + 64 = 134 \text{ g/mol} \]

Now, we can calculate the number of moles of \( Na_2C_2O_4 \):
\[ \text{Moles of } Na_2C_2O_4 = \frac{0.4 \text{ g}}{134 \text{ g/mol}} \approx 0.00299 \text{ mol} \]

### Step 4: Calculate the mass of \( UO_2(C_2O_4) \cdot xH_2O \)

We know that the reaction yields 1.23 g of \( UO_2(C_2O_4) \cdot xH_2O \). The molar mass of \( UO_2(C_2O_4) \) can be calculated as follows:
- Molar mass of \( UO_2 \) = 238 + 2(16) = 270 g/mol
- Molar mass of \( C_2O_4 \) = 2(12) + 4(16) = 88 g/mol

Thus, the molar mass of \( UO_2(C_2O_4) \):
\[ \text{Molar mass} = 270 + 88 + x(18) = 358 + 18x \text{ g/mol} \]

### Step 5: Set up the equation to find \( x \)

Using the number of moles calculated from the mass of \( UO_2(C_2O_4) \):
\[ \text{Moles of } UO_2(C_2O_4) = \frac{1.23 \text{ g}}{358 + 18x} \text{ mol} \]

Since \( Na_2C_2O_4 \) is the limiting reagent, we set the moles equal:
\[ 0.00299 = \frac{1.23}{358 + 18x} \]

### Step 6: Solve for \( x \)

Rearranging the equation:
\[ 1.23 = 0.00299(358 + 18x) \]
\[ 1.23 = 1.06942 + 0.05382x \]
\[ 1.23 - 1.06942 = 0.05382x \]
\[ 0.16058 = 0.05382x \]
\[ x \approx 2.98 \approx 3 \]

### Final Answers
1. Weight percentage of uranium in the original sample: **89.4%**
2. Value of \( x \) in \( UO_2(C_2O_4) \cdot xH_2O \): **3**