When sodium is heated with sulfur,sodium sulfide is product. What mass of sodium sulfide can be formed by heating 12.8 g of sulfur with 9.2g of sodium?

This is a limiting reagent problem; I know that because amounts are given for BOTH reactants. Here is a worked example of a limiting reagent problem. Just follow the steps. It's too dark to read on my screen; hope you have better luck. Let me know if you can't read it.

http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html

To determine the mass of sodium sulfide that can be formed, we first need to find the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.

We can find the limiting reactant by comparing the number of moles of sulfur and sodium with a balanced chemical equation:

2 Na + S → Na2S

The molar mass of sodium (Na) is 22.99 g/mol, and the molar mass of sulfur (S) is 32.07 g/mol.

First, we calculate the number of moles of sulfur:
moles of sulfur = mass / molar mass
moles of sulfur = 12.8 g / 32.07 g/mol = 0.399 mol

Next, we calculate the number of moles of sodium:
moles of sodium = mass / molar mass
moles of sodium = 9.2 g / 22.99 g/mol = 0.400 mol

According to the balanced equation, the ratio of sodium to sulfur is 2:1. This means that for every 2 moles of sodium, we need 1 mole of sulfur.

From the ratio, we can see that we have a 1:1 ratio of moles of sulfur and sodium. Therefore, the limiting reactant is sulfur.

Now we can calculate the mass of sodium sulfide formed.

The molar mass of sodium sulfide (Na2S) is:
molar mass of Na2S = (2 x molar mass of Na) + molar mass of S
molar mass of Na2S = (2 x 22.99 g/mol) + 32.07 g/mol
molar mass of Na2S = 45.98 g/mol + 32.07 g/mol
molar mass of Na2S = 78.05 g/mol

To calculate the mass of sodium sulfide formed, we use the following equation:

mass of sodium sulfide = moles of sulfur x molar mass of Na2S
mass of sodium sulfide = 0.399 mol x 78.05 g/mol
mass of sodium sulfide = 31.188 g

Therefore, the mass of sodium sulfide that can be formed by heating 12.8 g of sulfur with 9.2 g of sodium is approximately 31.188 grams.

To determine the mass of sodium sulfide that can be formed when 12.8 g of sulfur reacts with 9.2 g of sodium, we need to use the concept of stoichiometry.

1. Write the balanced chemical equation for the reaction:
2 Na + S → Na2S

2. Calculate the molar masses:
- Molar mass of sodium (Na) = 22.99 g/mol
- Molar mass of sulfur (S) = 32.06 g/mol
- Molar mass of sodium sulfide (Na2S) = 46.98 g/mol

3. Convert the given masses to moles:
- Moles of sulfur = mass / molar mass = 12.8 g / 32.06 g/mol = 0.399 mol
- Moles of sodium = mass / molar mass = 9.2 g / 22.99 g/mol = 0.400 mol

4. Determine the limiting reactant:
Compare the mole ratios of the reactants in the balanced equation. In this case, the ratio is 2:1 (2 moles of sodium for 1 mole of sulfur). Since the moles of sodium and sulfur are almost equal (0.400 vs. 0.399), the limiting reactant is sulfur because there is slightly less sulfur.

5. Calculate the moles of sodium sulfide formed:
According to the balanced equation, 1 mole of sulfur reacts to form 1 mole of sodium sulfide. Since sulfur is the limiting reactant, 0.399 moles of sulfur will form 0.399 moles of sodium sulfide.

6. Convert moles to grams:
Mass of sodium sulfide = moles of sodium sulfide × molar mass
Mass of sodium sulfide = 0.399 mol × 46.98 g/mol = 18.76 g

Therefore, approximately 18.76 grams of sodium sulfide can be formed by heating 12.8 g of sulfur with 9.2 g of sodium.