What mass in grams of sodium hydroxide is produced if 20.0 g of sodium metal reacts with excess water according to the chemical equation 2Na(s)+2H20(l)=2NaOH(aq)+H2(g)?
Please show me step by step how to do this because I have 5 other questions like this one on a worksheet and I have no idea how to do this stuff
Here are the steps to go through.
1. Write the balanced chemical equation.
2. Convert what you have into mols, in this case that is convert 20 g Na to mols remembering that mols = grams/molar mass (actually atomic mass in this case).
3. Now convert mols of what you have (Na here) to mols of what you want (in this case NaOH) using the coefficients in the balanced equation. As a big hint, you use dimensional analysis to do this.
4. Now convert mols of what you want (in this case mols NaOH from step 3) to grams. grams = mols x molar mass.
Let me know if you don't understand one of the parts but work as far as you can, post your work, and tell me exactly why you don't know the next step.
To solve this problem, you need to use stoichiometry, which is the calculation of the quantitative relationships between the reactants and products in a chemical equation. Here's how to solve it step by step:
Step 1: Write out the balanced chemical equation:
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
Step 2: Calculate the molar mass of sodium hydroxide (NaOH):
Na = 23 g/mol
O = 16 g/mol
H = 1 g/mol
NaOH = 23 + 16 + 1 = 40 g/mol
Step 3: Convert the given mass of sodium metal (Na) to moles:
20.0 g Na × (1 mol Na/23 g Na) = 0.8696 mol Na
Step 4: Use the stoichiometric ratio from the balanced equation to convert moles of sodium (Na) to moles of sodium hydroxide (NaOH):
0.8696 mol Na × (2 mol NaOH/2 mol Na) = 0.8696 mol NaOH
Step 5: Convert the moles of sodium hydroxide (NaOH) to grams:
0.8696 mol NaOH × (40 g NaOH/1 mol NaOH) = 34.8 g NaOH
Therefore, the mass of sodium hydroxide produced is 34.8 grams when 20.0 grams of sodium metal reacts with excess water.
To calculate the mass of sodium hydroxide produced in this reaction, follow these step-by-step instructions:
Step 1: Write down the balanced chemical equation.
The balanced equation is:
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
This equation tells us that 2 moles of sodium metal react with 2 moles of water to produce 2 moles of sodium hydroxide and 1 mole of hydrogen gas.
Step 2: Determine the molar mass of sodium hydroxide (NaOH).
The molar mass of sodium (Na) is 22.99 g/mol, the molar mass of oxygen (O) is 15.999 g/mol, and the molar mass of hydrogen (H) is 1.008 g/mol.
Add these values to get the molar mass of NaOH:
Molar mass of NaOH = (22.99 g/mol) + (15.999 g/mol) + (1.008 g/mol)
= 40.997 g/mol (round to 41.0 g/mol)
Step 3: Convert the given mass of sodium (Na) to moles.
Use the molar mass of sodium (22.99 g/mol) to convert the given mass of sodium (20.0 g) to moles.
Moles of Na = (Given mass of Na) / (Molar mass of Na)
= 20.0 g / 22.99 g/mol
≈ 0.870 moles of Na (rounded to three decimal places)
Step 4: Determine the limiting reactant.
To find the limiting reactant, compare the mole ratio of Na to NaOH from the balanced equation.
In the balanced equation, 2 moles of Na reacts with 2 moles of NaOH.
So for every 1 mole of Na, we expect to get 1 mole of NaOH.
Since we have 0.870 moles of Na, we will get 0.870 moles of NaOH.
Step 5: Calculate the mass of sodium hydroxide (NaOH) produced.
To convert moles of NaOH to grams, use the molar mass of NaOH (41.0 g/mol).
Mass of NaOH = (Moles of NaOH) × (Molar mass of NaOH)
= 0.870 moles × 41.0 g/mol
= 35.67 g (rounded to two decimal places)
Therefore, the mass of sodium hydroxide produced is approximately 35.67 grams.