If you start with 10.0 grams of lithium hydroxide (LiOH) how many grams of Ilthium bromide (LIBr) will be produced? LiOH+HBrarrow LBr+H_(2)O 7.58 758 2.76g 36.258 Multiple Choice 10 points

To determine how many grams of lithium bromide (LiBr) will be produced from 10.0 grams of lithium hydroxide (LiOH), we need to follow these steps:<br /><br />1. **Write the balanced chemical equation:**<br /> \[<br /> \text{LiOH} + \text{HBr} \rightarrow \text{LiBr} + \text{H}_2\text{O}<br /> \]<br /><br />2. **Calculate the molar masses of the reactants and products:**<br /> - Molar mass of LiOH: \( 6.94 \, (\text{Li}) + 15.999 \, (\text{O}) + 1.008 \, (\text{H}) = 23.946 \, \text{g/mol} \)<br /> - Molar mass of LiBr: \( 6.94 \, (\text{Li}) + 79.904 \, (\text{Br}) = 86.854 \, \text{g/mol} \)<br /><br />3. **Convert the mass of LiOH to moles:**<br /> \[<br /> \text{Moles of LiOH} = \frac{10.0 \, \text{g}}{23.946 \, \text{g/mol}} \approx 0.417 \, \text{mol}<br /> \]<br /><br />4. **Use the stoichiometry of the reaction to find the moles of LiBr produced:**<br /> According to the balanced equation, 1 mole of LiOH produces 1 mole of LiBr. Therefore, 0.417 moles of LiOH will produce 0.417 moles of LiBr.<br /><br />5. **Convert the moles of LiBr to grams:**<br /> \[<br /> \text{Mass of LiBr} = 0.417 \, \text{mol} \times 86.854 \, \text{g/mol} \approx 36.258 \, \text{g}<br /> \]<br /><br />Therefore, the correct answer is:<br />\[ \boxed{36.258 \, \text{g}} \]