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Answer :
Sure! Let's go through the solution step-by-step to find out how many grams of [tex]\( \text{Sr(OH)}_2 \)[/tex] were used in the reaction.
### Step 1: Understand the reaction
The chemical equation for the reaction is:
[tex]\[ \text{Sr(OH)}_2 + 2 \text{HF} \rightarrow \text{SrF}_2 + 2 \text{H}_2\text{O} \][/tex]
From the equation, we see that:
- 1 mole of [tex]\( \text{Sr(OH)}_2 \)[/tex] produces 1 mole of [tex]\( \text{SrF}_2 \)[/tex].
### Step 2: Determine moles of [tex]\( \text{Sr(OH)}_2 \)[/tex] used
Given that 2.00 moles of [tex]\( \text{SrF}_2 \)[/tex] are produced, and the stoichiometry from the balanced equation is 1:1, we also know:
- Moles of [tex]\( \text{Sr(OH)}_2 \)[/tex] used = Moles of [tex]\( \text{SrF}_2 \)[/tex] produced = 2.00 moles
### Step 3: Calculate the mass of [tex]\( \text{Sr(OH)}_2 \)[/tex]
The molar mass of [tex]\( \text{Sr(OH)}_2 \)[/tex] is approximately 121.632 g/mol. Using this, we can find the mass of [tex]\( \text{Sr(OH)}_2 \)[/tex] used.
To calculate the mass:
[tex]\[
\text{Mass of } \text{Sr(OH)}_2 = \text{moles of } \text{Sr(OH)}_2 \times \text{molar mass of } \text{Sr(OH)}_2
\][/tex]
[tex]\[
= 2.00 \, \text{moles} \times 121.632 \, \text{g/mol}
\][/tex]
[tex]\[
= 243.264 \, \text{grams}
\][/tex]
Therefore, 243.264 grams of [tex]\( \text{Sr(OH)}_2 \)[/tex] were used in the reaction.
### Step 1: Understand the reaction
The chemical equation for the reaction is:
[tex]\[ \text{Sr(OH)}_2 + 2 \text{HF} \rightarrow \text{SrF}_2 + 2 \text{H}_2\text{O} \][/tex]
From the equation, we see that:
- 1 mole of [tex]\( \text{Sr(OH)}_2 \)[/tex] produces 1 mole of [tex]\( \text{SrF}_2 \)[/tex].
### Step 2: Determine moles of [tex]\( \text{Sr(OH)}_2 \)[/tex] used
Given that 2.00 moles of [tex]\( \text{SrF}_2 \)[/tex] are produced, and the stoichiometry from the balanced equation is 1:1, we also know:
- Moles of [tex]\( \text{Sr(OH)}_2 \)[/tex] used = Moles of [tex]\( \text{SrF}_2 \)[/tex] produced = 2.00 moles
### Step 3: Calculate the mass of [tex]\( \text{Sr(OH)}_2 \)[/tex]
The molar mass of [tex]\( \text{Sr(OH)}_2 \)[/tex] is approximately 121.632 g/mol. Using this, we can find the mass of [tex]\( \text{Sr(OH)}_2 \)[/tex] used.
To calculate the mass:
[tex]\[
\text{Mass of } \text{Sr(OH)}_2 = \text{moles of } \text{Sr(OH)}_2 \times \text{molar mass of } \text{Sr(OH)}_2
\][/tex]
[tex]\[
= 2.00 \, \text{moles} \times 121.632 \, \text{g/mol}
\][/tex]
[tex]\[
= 243.264 \, \text{grams}
\][/tex]
Therefore, 243.264 grams of [tex]\( \text{Sr(OH)}_2 \)[/tex] were used in the reaction.
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