U(OH)6+3SmO-gt UO3+3Sm(OH)2 How many moles of UO3 do I get from 5.11 moles of U(OH)6 and 7.89 moles of Smo 5.11 7.89 2.63 1.70

To determine how many moles of UO3 are produced, we need to use stoichiometry based on the balanced chemical equation:<br /><br />\[ U(OH)_6 + 3SmO \rightarrow UO_3 + 3Sm(OH)_2 \]<br /><br />From the balanced equation, we see that 1 mole of \( U(OH)_6 \) reacts with 3 moles of \( SmO \) to produce 1 mole of \( UO_3 \).<br /><br />First, let's find the limiting reagent:<br /><br />1. Calculate the required moles of \( SmO \) for 5.11 moles of \( U(OH)_6 \):<br /><br />\[ 5.11 \text{ moles } U(OH)_6 \times \frac{3 \text{ moles } SmO}{1 \text{ mole } U(OH)_6} = 15.33 \text{ moles } SmO \]<br /><br />2. Compare the available moles of \( SmO \) (7.89 moles) with the required moles (15.33 moles):<br /><br />Since we only have 7.89 moles of \( SmO \), which is less than the required 15.33 moles, \( SmO \) is the limiting reagent.<br /><br />3. Calculate the moles of \( UO_3 \) produced from the limiting reagent (\( SmO \)):<br /><br />\[ 7.89 \text{ moles } SmO \times \frac{1 \text{ mole } UO_3}{3 \text{ moles } SmO} = 2.63 \text{ moles } UO_3 \]<br /><br />Therefore, you get 2.63 moles of \( UO_3 \).