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Consider the balanced chemical equation:

[tex]\[ 3 \text{Sr(OH)}_2(aq) + 2 \text{H}_3\text{PO}_4(aq) \rightarrow \text{Sr}_3(\text{PO}_4)_2(s) + 6 \text{H}_2\text{O}(l) \][/tex]

How many moles of [tex]\(\text{Sr(OH)}_2\)[/tex] remain if 0.10 mol of [tex]\(\text{H}_3\text{PO}_4\)[/tex] react with 0.40 mol of [tex]\(\text{Sr(OH)}_2\)[/tex]?

A. 0.30 mol
B. 0.050 mol
C. 0.20 mol
D. 0.25 mol

Sagot :

GinnyAnsweridn
Let's solve this problem step-by-step.

### Step 1: Understand the Balanced Chemical Equation
The balanced chemical equation is:
[tex]\[ 3 \text{Sr(OH)}_2 (aq) + 2 \text{H}_3\text{PO}_4 (aq) \rightarrow \text{Sr}_3(\text{PO}_4)_2 (s) + 6 \text{H}_2\text{O} (\ell) \][/tex]

From the equation, the stoichiometric coefficients are:
- 3 moles of Sr(OH)[tex]\(_2\)[/tex] react with 2 moles of H[tex]\(_3\)[/tex]PO[tex]\(_4\)[/tex].

### Step 2: Given Quantities
The problem states that we have:
- 0.40 moles of Sr(OH)[tex]\(_2\)[/tex]
- 0.10 moles of H[tex]\(_3\)[/tex]PO[tex]\(_4\)[/tex]

### Step 3: Determine the Limiting Reactant
We must determine which reactant will be completely consumed first (the limiting reactant).

Calculate the moles of Sr(OH)[tex]\(_2\)[/tex] required to completely react with the given moles of H[tex]\(_3\)[/tex]PO[tex]\(_4\)[/tex]:
[tex]\[ \text{required Sr(OH)}_2 = \frac{3 \ \text{moles of Sr(OH)}_2}{2 \ \text{moles of H}_3\text{PO}_4} \times \text{moles of H}_3\text{PO}_4 \][/tex]
[tex]\[ \text{required Sr(OH)}_2 = \frac{3}{2} \times 0.10 \][/tex]
[tex]\[ \text{required Sr(OH)}_2 = 0.15 \ \text{moles} \][/tex]

The calculation shows that 0.15 moles of Sr(OH)[tex]\(_2\)[/tex] are required to fully react with 0.10 moles of H[tex]\(_3\)[/tex]PO[tex]\(_4\)[/tex].

### Step 4: Compare Available and Required Moles
We have 0.40 moles of Sr(OH)[tex]\(_2\)[/tex] available and only 0.15 moles are needed to react with the H[tex]\(_3\)[/tex]PO[tex]\(_4\)[/tex].

Since [tex]\( \text{available Sr(OH)}_2 (0.40) > \text{required Sr(OH)}_2 (0.15) \)[/tex]:

- H[tex]\(_3\)[/tex]PO[tex]\(_4\)[/tex] is the limiting reactant.
- Some Sr(OH)[tex]\(_2\)[/tex] will be left unreacted.

### Step 5: Calculate Remaining Sr(OH)[tex]\(_2\)[/tex]
The remaining moles of Sr(OH)[tex]\(_2\)[/tex] can be calculated as follows:
[tex]\[ \text{remaining Sr(OH)}_2 = \text{initial Sr(OH)}_2 - \text{required Sr(OH)}_2 \][/tex]
[tex]\[ \text{remaining Sr(OH)}_2 = 0.40 - 0.15 \][/tex]
[tex]\[ \text{remaining Sr(OH)}_2 = 0.25 \][/tex]

### Step 6: Conclusion
The moles of Sr(OH)[tex]\(_2\)[/tex] that remain unreacted are:
[tex]\[ 0.25 \ \text{moles} \][/tex]

Thus, the correct answer is:
[tex]\[ \boxed{0.25 \ \text{mol}} \][/tex]

The correct choice is: D. 0.25 mol
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