Answer:
A) NO₂ is in excess and 2.5 moles NO₂ remain after the reaction has completed.
B) 45.5 ml of the 0.170 Na₃PO₄ solution is necessary to completely react with 94.8 ml of 0.122 M CuCl₂
Step-by-step explanation:
Hi there!
A) From the equation, we know that 1 mol H₂O reacts with 3 mol NO₂. Then, 0.40 mol H₂O will react with (0.40 mol H₂O · (3 mol NO₂ / 1 mol H₂O)) 1.2 mol NO₂.
Since we have 3.7 mol NO₂ and only 1.2 mol will react, we have (3.7-1.2) 2.5 mol NO₂ in excess.
B) From the equation, we know that 2 mol Na₃PO₄ reacts with 3 mol CuCl₂.
First, let´s calculate how many moles of CuCl₂ are present in 94.8 ml of a 0.122 M solution:
94.8 ml · (0.122 mol CuCl₂ / 1000 ml) = 0.0116 mol CuCl₂
Since 3 mol CuCl₂ react with 2 mol Na₃PO₄, then, 0.0116 mol CuCl₂ will react with (0.0116 mol CuCl₂ · (2 mol Na₃PO₄ / 3 mol CuCl₂)) 7.73 × 10⁻³ mol Na₃PO₄
Now, let´s find the volume of the 0.170 M solution that contains that amount of moles of Na₃PO₄
We know that 0.170 moles are present in 1000 ml of the solution. Then, 7.73 × 10⁻³ mol will be present in (7.73 × 10⁻³ mol · (1000 ml / 0.170 mol)) 45.5 ml.
45.5 ml of the 0.170 Na₃PO₄ solution is necessary to completely react with 94.8 ml of 0.122 M CuCl₂