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Answer:
The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.
Explanation:
The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:
n = 1.375x10⁻⁵ mol
The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):
1.375x10⁻⁵ mol _________ 19.9 units
x _________ 25.2 units
x = 1.741x10⁻⁵mol
Finally, we can calculate the Cu²⁺ concentration :
C = 1.741x10⁻⁵mol / 0.025 L
C = 6.964x10⁻⁴ M
Answer:
NaOH is the limiting reactant.
204.9 g of sodium phosphate are formed.
51.94 g of excess reactant will remain.
Explanation:
The reaction that takes place is:
- H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O
First we <u>convert the mass of both reactants to moles</u>, using their <em>respective molar masses</em>:
- H₃PO₄ ⇒ 175 g ÷ 98 g/mol = 1.78 mol
- NaOH ⇒ 150 g ÷ 40 g/mol = 3.75 mol
1.78 moles of H₃PO₄ would react completely with (1.78 * 3) 5.34 moles of NaOH. There are not as many NaOH moles so NaOH is the limiting reactant.
--
We <u>calculate the produced moles of Na₃PO₄</u> using the <em>limiting reactant</em>:
- 3.75 mol NaOH *
= 1.25 mol Na₃PO₄
Then we <u>convert moles into grams</u>:
- 1.25 mol Na₃PO₄ * 163.94 g/mol = 204.9 g
--
We calculate how many H₃PO₄ moles would react with 3.75 NaOH moles:
- 3.75 mol NaOH *
= 1.25 mol H₃PO₄
We substract that amount from the original amount:
- 1.78 - 1.25 = 0.53 mol H₃PO₄
Finally we <u>convert those remaining moles to grams</u>:
- 0.53 mol H₃PO₄ * 98 g/mol = 51.94 g