C₄H₉OH + HBr = C₄H₉Br + H2O
Δmole of alcohol gives 1 mole of bromobutanol
HBr is in excess, so the yield of the product is limited by the alcohol
Wt. of 1 butanol = 18
Molar mass of the butanol = 74.12 g/mole
Moles of the alcohol = 1/74.12 = 0.01349 moles
So, moles of bromobutane = 0.01349 moles
Molar mass of C₄H₉Br = 137.018 g/moles
So, theoretical mass of bromobutane is = 0.01349 × 137.0.18
= 1.85 g
Answer:
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Explanation:
Answer:
Explanation:
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In this case, since we are considering an gas, which can be considered as idea, we can write the ideal gas equation in order to write it in terms of density rather than moles and volume:
Whereas MM is the molar mass of the gas. Now, since we can identify the initial and final states, we can cancel out R and MM since they remain the same:
It means we can compute the final density as shown below:
Now, we plug in to obtain:
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