madey21
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[tex]\[ 2 \text{HNO}_3 + \text{Ba(OH)}_2 \rightarrow 2 \text{H}_2\text{O} + \text{Ba}\left(\text{NO}_3\right)_2 \][/tex]

50.0 mL of 0.250 M [tex]\(\text{HNO}_3\)[/tex] is added to 30.0 mL of 0.400 M [tex]\(\text{Ba(OH)}_2\)[/tex]. How many moles of [tex]\(\text{Ba}\left(\text{NO}_3\right)_2\)[/tex] could form from the [tex]\(\text{HNO}_3\)[/tex]?

The [tex]\(\text{HNO}_3\)[/tex] forms [tex]\(\square\)[/tex] mol [tex]\(\text{Ba}\left(\text{NO}_3\right)_2\)[/tex].

Sagot :

GinnyAnswer
Certainly! To determine how many moles of [tex]\( \text{Ba(NO}_3\text{)}_2 \)[/tex] are produced from the given quantities of [tex]\( \text{HNO}_3 \)[/tex], we will follow these steps:

1. Write the balanced chemical equation:
[tex]\[ 2 \text{HNO}_3 + \text{Ba(OH)}_2 \rightarrow 2 \text{H}_2\text{O} + \text{Ba(NO}_3\text{)}_2 \][/tex]

2. Determine the moles of [tex]\( \text{HNO}_3 \)[/tex] and [tex]\( \text{Ba(OH)}_2 \)[/tex] we start with:
- Volume of [tex]\( \text{HNO}_3 = 50.0 \)[/tex] mL
- Concentration of [tex]\( \text{HNO}_3 = 0.250 \)[/tex] M
- Volume of [tex]\( \text{Ba(OH)}_2 = 30.0 \)[/tex] mL
- Concentration of [tex]\( \text{Ba(OH)}_2 = 0.400 \)[/tex] M

3. Convert volumes from mL to L:
- Volume [tex]\( \text{HNO}_3 \)[/tex] in liters: [tex]\( \frac{50.0}{1000} = 0.050 \, \text{L} \)[/tex]
- Volume [tex]\( \text{Ba(OH)}_2 \)[/tex] in liters: [tex]\( \frac{30.0}{1000} = 0.030 \, \text{L} \)[/tex]

4. Calculate the moles of each reactant:
- Moles of [tex]\( \text{HNO}_3 \)[/tex] = Volume [tex]\( \times \)[/tex] Concentration = [tex]\( 0.050 \, \text{L} \times 0.250 \, \text{M} = 0.0125 \, \text{mol} \)[/tex]
- Moles of [tex]\( \text{Ba(OH)}_2 \)[/tex] = Volume [tex]\( \times \)[/tex] Concentration = [tex]\( 0.030 \, \text{L} \times 0.400 \, \text{M} = 0.012 \, \text{mol} \)[/tex]

5. Identify the limiting reagent:
According to the balanced equation, 2 moles of [tex]\( \text{HNO}_3 \)[/tex] react with 1 mole of [tex]\( \text{Ba(OH)}_2 \)[/tex]. Hence, the stoichiometric ratio is 2:1.

Compare the ratio of the moles you have:
- Ratio of [tex]\( \text{HNO}_3 \)[/tex] to [tex]\( \text{Ba(OH)}_2 \)[/tex] is [tex]\( 0.0125 \, \text{mol} : 0.012 \, \text{mol} \)[/tex].
- To satisfy the reaction's stoichiometry (2:1), 0.0125 moles of [tex]\( \text{HNO}_3 \)[/tex] would require [tex]\( 0.0125 / 2 = 0.00625 \, \text{mol} \)[/tex] of [tex]\( \text{Ba(OH)}_2 \)[/tex].

Since we have only 0.012 moles of [tex]\( \text{Ba(OH)}_2 \)[/tex], the [tex]\( \text{HNO}_3 \)[/tex] is the limiting reagent.

6. Calculate the moles of [tex]\( \text{Ba(NO}_3\text{)}_2 \)[/tex] that can form:
From the balanced equation, 2 moles of [tex]\( \text{HNO}_3 \)[/tex] produce 1 mole of [tex]\( \text{Ba(NO}_3\text{)}_2 \)[/tex].

So, 0.0125 moles of [tex]\( \text{HNO}_3 \)[/tex] will produce:
[tex]\[ \text{Moles of Ba(NO}_3\text{)}_2 = \frac{0.0125 \, \text{mol}}{2} = 0.00625 \, \text{mol} \][/tex]

Therefore, the [tex]\( \text{HNO}_3 \)[/tex] forms [tex]\( 0.00625 \)[/tex] mol [tex]\( \text{Ba(NO}_3\text{)}_2 \)[/tex].
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