1.80 g of a metal oxide required 833 mL of hydrogen at NTP to be reduced to its metal. Find the equivalent weights of the oxide and the metal.

Explanation

To find the equivalent weights of the metal oxide and the metal, we can follow these steps:

### Step 1: Calculate the moles of hydrogen gas used
We know that at Normal Temperature and Pressure (NTP), 1 mole of gas occupies 22.4 L. First, we need to convert the volume of hydrogen gas from mL to L.

\[ \text{Volume of H}_2 = 833 \, \text{mL} = \frac{833}{1000} \, \text{L} = 0.833 \, \text{L} \]
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