What is the theoretical yield of Na2SO4 when 10 g of Na reacts with excess sulfuric acid?
Practice Questions
1 question
Q1
What is the theoretical yield of Na2SO4 when 10 g of Na reacts with excess sulfuric acid?
20 g
30 g
40 g
50 g
10 g of Na = 0.434 moles. 2Na + H2SO4 → Na2SO4 + H2. 0.434 moles of Na2SO4 = 0.434 * 142 g = 61.6 g.
Questions & Step-by-step Solutions
1 item
Q
Q: What is the theoretical yield of Na2SO4 when 10 g of Na reacts with excess sulfuric acid?
Solution: 10 g of Na = 0.434 moles. 2Na + H2SO4 → Na2SO4 + H2. 0.434 moles of Na2SO4 = 0.434 * 142 g = 61.6 g.
Steps: 4
Step 1: Calculate the number of moles of sodium (Na) in 10 grams. Use the formula: moles = mass (g) / molar mass (g/mol). The molar mass of Na is approximately 23 g/mol. So, moles of Na = 10 g / 23 g/mol = 0.434 moles.
Step 2: Write the balanced chemical equation for the reaction between sodium (Na) and sulfuric acid (H2SO4). The equation is: 2Na + H2SO4 → Na2SO4 + H2.
Step 3: From the balanced equation, notice that 2 moles of Na produce 1 mole of Na2SO4. Therefore, 0.434 moles of Na will produce 0.434 / 2 = 0.217 moles of Na2SO4.
Step 4: Calculate the mass of Na2SO4 produced using the number of moles. The molar mass of Na2SO4 is approximately 142 g/mol. So, mass of Na2SO4 = moles × molar mass = 0.217 moles × 142 g/mol = 30.8 g.
