Question

How many moles of Sodium sulfide are needed if 20 g of Sodium nitrate have to be produced?

a) 0.25 moles
b) 0.50 moles
c) 0.75 moles
d) 1.0 moles

Answer 1

The number of moles of Sodium sulfide needed to produce 20 g of Sodium nitrate is 0.12 moles.

Step-by-step explanation:

To determine the number of moles of Sodium sulfide needed to produce 20 g of Sodium nitrate, we need to use the stoichiometry of the balanced chemical equation and the molar masses of the compounds. The balanced equation for the reaction is:

2 Na₂S + 4 NaNO₃ → 6 Na⁺ + 2 N₂ + 4 O₂ + 2 S

The molar mass of NaNO₃ is 85 g/mol. Therefore, the number of moles of NaNO₃ is 20 g / 85 g/mol = 0.235 moles.

From the balanced equation, we can see that 4 moles of NaNO₃ are needed to produce 2 moles of Na₂S.

So, the number of moles of Na₂S needed is (0.235 moles / 4 moles) x 2 moles = 0.1175 moles.

Rounding to the nearest hundredth, the answer is 0.12 moles.