How many moles of HNO3 can react with 3.8g of Ca(OH)2?

\textbf{Step 1: Balanced Chemical Equation}

The balanced chemical equation for the reaction between Ca(OH)_2 and HNO_3 is:

Ca(OH)_2 + 2HNO_3 \rightarrow Ca(NO_3)_2 + 2H_2O

\textbf{Step 2: Calculate Molar Mass of } Ca(OH)_2

The molar mass of Ca(OH)_2 is calculated as:

Molar\ Mass\ of\ Ca(OH)_2 = 40.08 + (2 \times 16.00) + (2 \times 1.01) = 74.1\ g/mol

\textbf{Step 3: Calculate Moles of } Ca(OH)_2

The number of moles of Ca(OH)_2 present in 3.8 grams is calculated as:

Moles\ of\ Ca(OH)_2 = \frac{3.8\ g}{74.1\ g/mol} \approx 0.0513\ moles

\textbf{Step 4: Determine Moles of } HNO_3 \textbf{ that React with } Ca(OH)_2

Since 2 moles of HNO3 react with 1 mole of Ca(OH)2, we calculate the moles of HNO3 needed:

Moles\ of\ HNO_3 = 2 \times 0.0513\ moles \approx 0.1026\ moles

\textbf{Answer:} About 0.1026 moles of HNO_3 can react with 3.8 grams of Ca(OH)_2 .