2Na+2H2O-->2NaOH+H2
what is the mass of reactant left over?
pls show work too thank you
To find the mass of the reactant left over, we need to determine which reactant is the limiting reactant and then calculate the amount of excess reactant.
1. Start by identifying the limiting reactant. In this reaction, we have two reactants: sodium (Na) and water (H2O). To do this, we need to compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation.
The stoichiometric ratio for the reaction is 2 moles of sodium (Na) to 2 moles of water (H2O), meaning they react in a 1:1 ratio.
2. Convert the mass of each reactant to moles.
The molar mass of sodium (Na) is 22.99 g/mol.
The molar mass of water (H2O) is 18.02 g/mol.
Let's assume we have 10 grams of sodium (Na) and 20 grams of water (H2O).
Moles of sodium (Na) = 10 g Na * (1 mol Na / 22.99 g Na)
= 0.435 mol Na
Moles of water (H2O) = 20 g H2O * (1 mol H2O / 18.02 g H2O)
= 1.11 mol H2O
3. Compare the moles of reactants based on the stoichiometric ratio.
The stoichiometric ratio is 1:1 for Na:H2O.
Since we have 0.435 mol Na and 1.11 mol H2O, the Na is the limiting reactant because we have less moles of Na than H2O.
4. Calculate the moles of the product formed based on the limiting reactant.
According to the balanced equation, the reaction produces 2 moles of NaOH for every 2 moles of Na. Therefore, the moles of NaOH formed are also 0.435 mol.
5. Calculate the moles of the excess reactant that reacted.
Moles of excess H2O = Moles of reactant initially used - Moles of product formed
= 1.11 mol H2O - 0.435 mol NaOH
= 0.675 mol H2O
6. Convert the moles of the excess reactant to mass.
Mass of excess H2O = Moles of excess H2O * Molar mass of H2O
= 0.675 mol H2O * (18.02 g H2O / 1 mol H2O)
= 12.15 g H2O
Therefore, the mass of the reactant left over is 12.15 grams of water (H2O).
To find the mass of reactant left over, we need to follow these steps:
Step 1: Calculate the molar mass of Na (sodium) and H2O (water).
The molar mass of Na = 22.99 g/mol
The molar mass of H2O = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
Step 2: Convert the given mass of Na and H2O into moles.
Let's assume we have 10 grams of Na and 20 grams of H2O.
Number of moles of Na = mass of Na / molar mass of Na
Number of moles of Na = 10 g / 22.99 g/mol ≈ 0.435 mol
Number of moles of H2O = mass of H2O / molar mass of H2O
Number of moles of H2O = 20 g / 18.02 g/mol ≈ 1.110 mol
Step 3: Determine the limiting reactant.
The balanced equation tells us that the stoichiometric ratio between Na and H2O is 2:2 (or 1:1). Therefore, the reactant that produces the least amount of product is the limiting reactant.
In this case, both Na and H2O have a stoichiometric coefficient of 2, indicating that both the reactants are fully utilized, and there is no excess reactant left over.
Hence, the mass of reactant left over is 0 grams for both Na and H2O.