Solution The conversion steps are grams of H_(2)arrow moles of H_(2)arrow moles of L_(1)arrow grams c Combining these steps into one equation, we write 9.89gH_(2)times (1molH_(2))/(2.016gH_(2))times (2molH_(2))/(1molH_(2))times (6.941gLi)/(1molH_(2))=68.1gLi Check There are roughly 5 moles of H_(2) in 9.89 g H_(2) so we need 10 moles of Li. From the approximate molar mass of Li (7 g), does the answer seem reasonable?

Question

Solution The conversion steps are grams of H_(2)arrow moles of H_(2)arrow moles of L_(1)arrow grams c Combining these steps into one equation, we write 9.89gH_(2)times (1molH_(2))/(2.016gH_(2))times (2molH_(2))/(1molH_(2))times (6.941gLi)/(1molH_(2))=68.1gLi Check There are roughly 5 moles of H_(2) in 9.89 g H_(2) so we need 10 moles of Li. From the approximate molar mass of Li (7 g), does the answer seem reasonable?

Answer 1

The correct answer is:68.1 g LiExplanation:The given problem involves converting grams of hydrogen gas (H2) to grams of lithium (Li). The conversion steps are:1. Convert grams of H2 to moles of H2 using the molar mass of H2 (2.016 g/mol).2. Convert moles of H2 to moles of Li using the stoichiometric factor of 2 mol H2/1 mol Li.3. Convert moles of Li to grams of Li using the molar mass of Li (6.941 g/mol).By multiplying the given mass of H2 (9.89 g) by the appropriate conversion factors, we can calculate the final mass of Li, which is 68.1 g. This answer seems reasonable, as the molar mass of Li is approximately 7 g/mol, and the problem states that there are roughly 5 moles of H2 in 9.89 g of H2, which would require 10 moles of Li.

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