QQuestionChemistry
QuestionChemistry
Calculate the mass, in grams, of Al(2$)3 required for this reaction:
Reactants:
Al(2$)3
Products:
Al^2O^3 + H^2O
Given masses:
Al^2O^3 = 21.8 grams
H^2O = 9.7 grams
Enter only a number for your answer.
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Answer
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Step 1:I'll solve this step by step using precise LaTeX formatting:
Step 2:: Balance the chemical equation
2 \text{Al(OH)}_{3} \rightarrow \text{Al}_{2}\text{O}_{3} + 3\text{H}_{2}\text{O}
First, let's balance the chemical equation:
Step 3:: Calculate molar masses
- $$\text{Al}_{2}\text{O}_{3}: 2(26.98) + 3(16.00) = 101.96 \text{ g/mol}
Molar masses:
Step 4:: Calculate moles of Al^2O^3
n_{\text{Al}_{2}\text{O}_{3}} = \frac{21.8 \text{ g}}{101.96 \text{ g/mol}} = 0.214 \text{ mol}
Step 5:: Use stoichiometric ratio
From balanced equation, $$\frac{2 \text{ mol Al(OH)}_{3}}{1 \text{ mol Al}_{2}\text{O}_{3}}
Step 6:: Calculate moles of Al(OH)3
n_{\text{Al(OH)}_{3}} = 0.214 \text{ mol} \times \frac{2}{1} = 0.428 \text{ mol}
Step 7:: Calculate mass of Al(OH)3
m_{\text{Al(OH)}_{3}} = 0.428 \text{ mol} \times 78.00 \text{ g/mol} = 33.4
Final Answer
33.4
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