2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
c₁=2.00 mol/L
v₁=0.25 L
v₂=2.00 L
c₂-?
n(NaOH)=c₂v₂
n(H₂SO₄)=c₁v₁
n(NaOH)=2n(H₂SO₄)
c₂v₂=2c₁v₁
c₂=2c₁v₁/v₂
c₂=2*2.00*0.25/2.00=0.5 mol/L
0.5 M NaOH
18g is the most reasonable mass after the reaction
Based on the balanced chemical reaction presented above, every mole of magnesium (Mg) yields one mole of diatomic hydrogen (H2). When converted to masses, every 24.3 grams of magnesium yields 2 grams of hydrogen.
From the given, there are 20 grams of magnesium available for the reaction. With this amount, the expected yield of hydrogen is 1.646 grams. To calculate the percent yield, divide the actual yield to the hypothetical yield.
*The case is impossible because the actual yield is greater than the theoretical yield.
If we assume that there had been a typographical error and that the actual yield is 0.7 grams instead of 1.7 grams, the percent yield becomes 42.5%. Thus, the answer is letter E.
