a) why should you use moles in stoichimetric problems?

b) solid reacts violently with water, producing heat, hydrogen gas, and sodium hydroxide. how many molecules of hydrogen gas are formed when 48.7 g of soodium are added to water?
2Na + 2H2P ------> 2NaOH + H2

Answered above.

a) Mole units are used in stoichiometric problems because they provide a convenient way to represent the quantities of different substances involved in a chemical reaction. Mole is a counting unit that represents a specific number of particles, which allows chemists to compare and calculate the amounts of reactants and products in a chemical equation. By using moles, we can easily determine the ratios of different substances involved in a reaction and perform calculations such as determining the limiting reactant or predicting the amount of product formed.

b) To solve this stoichiometric problem, we need to use the given equation to determine the molar ratio between sodium (Na) and hydrogen gas (H2), and then use the molar mass of sodium to convert grams to moles. Finally, we can use the stoichiometry of the reaction to find the number of moles of hydrogen gas produced, and then convert moles to molecules using Avogadro's number.

1. The balanced equation is: 2Na + 2H2O → 2NaOH + H2

2. Calculate the molar mass of sodium (Na) using the periodic table. It is found to be 22.99 g/mol.

3. Convert grams of sodium (Na) to moles by dividing the given mass (48.7 g) by the molar mass of sodium (22.99 g/mol):
moles of Na = 48.7 g Na / 22.99 g/mol = 2.12 mol Na

4. Using the balanced equation, we can see that the mole ratio between sodium (Na) and hydrogen gas (H2) is 2 moles of Na : 1 mole of H2.

5. Apply the mole ratio to find the number of moles of hydrogen gas produced:
moles of H2 = 2 moles Na x (1 mole H2 / 2 moles Na) = 1 mol H2

6. Finally, convert moles of hydrogen gas to molecules using Avogadro's number, which states that there are 6.022 x 10^23 molecules in 1 mole:
molecules of H2 = 1 mol H2 x (6.022 x 10^23 molecules/mol) = 6.022 x 10^23 molecules of H2

Therefore, when 48.7 g of sodium reacts with water, it produces approximately 6.022 x 10^23 molecules of hydrogen gas.