Hey there !
Given the reaction:
N2 + 3 H2 = 2 NH3
At constant pressure and temperature ,volume is proporcional to moles:
Theoretical moles of N2 and H2 => 1:3
Theoretical volume of N2 and H2 => 1:3
Experimental volume of N2 and H2 => 3.0 L : 4.0 L
0.75 : 1 = 2.25 : 3
Since N2 is in excess reactant , H2 is the limiting reactant
Therefore:
volume of NH3 is 2/3 * Volume of H2
= 2/3 * 4.0 = 2.66 L
Hope that helps!
Molecular formula: C10H15Cl5
The balanced equation between NaOH and H₂SO₄ is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of moles of NaOH moles reacted = molarity of NaOH x volume
number of NaOH moles = 0.08964 mol/L x 27.86 x 10⁻³ L = 2.497 x 10⁻³ mol
according to molar ratio of 2:1
2 mol of NaOH reacts with 1 mol of H₂SO₄
therefore 2.497 x 10⁻³ mol of NaOH reacts with - 1/2 x 2.497 x 10⁻³ mol of H₂SO₄
number of moles of H₂SO₄ reacted - 1.249 x 10⁻³ mol
Number of H₂SO₄ moles in 34.53 mL - 1.249 x 10⁻³ mol
number of H₂SO₄ moles in 1000 mL - 1.249 x 10⁻³ mol / 34.53 x 10⁻³ L = 0.03617 mol
molarity of H₂SO₄ is 0.03617 M
As far as I know, the answer is B (breaking a rock)
