$$CaH_2 + 2H_2O → Ca(OH)_2 + 2H_2$$

Given:

- Mass of hydrogen gas formed = 4.850 g

Calculations:

1. First, calculate the moles of hydrogen gas produced.

$$Moles \ of \ H_2 = 4.850 \ g \ H_2 \times \frac{1 \ mol \ H_2}{2.016 \ g \ H_2} = 2.406 \ mol \ H_2$$

2. According to the balanced chemical equation, 1 mole of calcium hydride produces 2 moles of hydrogen gas. So, the moles of calcium hydride required can be calculated as follows:

$$Moles \ of \ CaH_2 = 2.406 \ mol \ H_2 \times \frac{1 \ mol \ CaH_2}{2 \ mol \ H_2} = 1.203 \ mol \ CaH_2$$

3. Finally, convert the moles of calcium hydride back to grams using its molar mass (42.09 g/mol):

$$Mass \ of \ CaH_2 = 1.203 \ mol \ CaH_2 \times 42.09 \ g/mol = 50.64 \ g \ CaH_2$$

Conclusion:

Approximately 50.64 grams of calcium hydride are required to produce 4.850 grams of hydrogen gas.