Question

how many ml of 3.000 m mg(oh)₂ solution are needed to react with 2.000 l of 3.000 m h₃po₄?
2 h₃po₄ + 3 mg(oh)₂ → 6 h₂o + mg₃(po₄)₂
this is a titration problem, and we are given enough information to calculate the moles of the acid.
3.000 mol
(2.000 l)(---------) = 1 mol h₃po₄
1 l
2 mol mg(oh)₂
(3 mol h₃po₄)(-------------------) = 4 mol mg(oh)₂
5 mol h₃po₄
1 l
(6 mol mg(oh)₂)(----------------) = 7 l mg(oh)₂
8 mol mg(oh)₂
9 ml
(10 l naoh)(----------) = 3.000 × 10³ ml mg(oh)₂
1 l
a. 3.000 b. 2.500 c. 10.00 d. 1.000
e. 2.000 f. 9.000 g. 6.000 h. 8.000 i. 0.02500
j. 47.00 k. 5.00 l. 50.00 m. 25.0 n. 0.02041
o. 97.98 p. 30.62 q. 6.572 r. 0.06124 s. 262.87

Explanation:

Step1: Calculate moles of \(H_3PO_4\)

Use the formula \(n = C\times V\), where \(C = 3.000\space M\) (molarity) and \(V=2.000\space L\) (volume).
\((2.000\space L)\times\frac{3.000\space mol}{1\space L}=6.000\space mol\space H_3PO_4\). So, blank 1 is \(6.000\) (matching option G).

Step2: Use mole - ratio from the balanced equation

From \(2H_3PO_4+3Mg(OH)_2
ightarrow6H_2O + Mg_3(PO_4)_2\), the mole - ratio of \(Mg(OH)_2\) to \(H_3PO_4\) is \(\frac{3}{2}\). So, blank 2 is \(3\), blank 5 is \(2\).
\((6.000\space mol\space H_3PO_4)\times\frac{3\space mol\space Mg(OH)_2}{2\space mol\space H_3PO_4}=9.000\space mol\space Mg(OH)_2\). So, blank 3 is \(6.000\), blank 4 is \(9.000\) (matching option F).

Step3: Calculate volume of \(Mg(OH)_2\) solution

Use \(V=\frac{n}{C}\), where \(n = 9.000\space mol\) and \(C = 3.000\space M\).
\((9.000\space mol\space Mg(OH)_2)\times\frac{1\space L}{3.000\space mol\space Mg(OH)_2}=3.000\space L\space Mg(OH)_2\). So, blank 6 is \(9.000\), blank 8 is \(3.000\) (matching option A).

Step4: Convert liters to milliliters

Use the conversion factor \(1\space L=1000\space mL\).
\((3.000\space L\space Mg(OH)_2)\times\frac{1000\space mL}{1\space L}=3000\space mL\space Mg(OH)_2\). So, blank 9 is \(1000\), blank 10 is \(3.000\).

Answer:

1. G. \(6.000\)
2. A. \(3.000\)
3. G. \(6.000\)
4. F. \(9.000\)
5. E. \(2.000\)
6. F. \(9.000\)
7. A. \(3.000\)
8. A. \(3.000\)
9. C. \(10.00\)
10. A. \(3.000\)