Bh2S7+7FmCO3-gt BH2(CO3)7+7 FmS How many moles of Fms do I get from 0.48 mol Bh2S7 and 291 mol FmCO3 ? 0.416 0.069 2.91 3.36

To determine the number of moles of FmS produced from 0.48 mol of Bh2S7 and 291 mol of FmCO3, we need to use stoichiometry based on the balanced chemical equation:<br /><br />\[ Bh2S7 + 7FmCO3 \rightarrow Bh2(CO3)7 + 7FmS \]<br /><br />From the equation, we see that 1 mole of Bh2S7 reacts with 7 moles of FmCO3 to produce 7 moles of FmS.<br /><br />First, let's find out how many moles of FmCO3 are needed to react with 0.48 moles of Bh2S7:<br /><br />\[ 0.48 \text{ mol Bh2S7} \times \frac{7 \text{ mol FmCO3}}{1 \text{ mol Bh2S7}} = 3.36 \text{ mol FmCO3} \]<br /><br />Next, we need to determine how many moles of FmS are produced from 3.36 moles of FmCO3:<br /><br />\[ 3.36 \text{ mol FmCO3} \times \frac{7 \text{ mol FmS}}{7 \text{ mol FmCO3}} = 3.36 \text{ mol FmS} \]<br /><br />Therefore, from 0.48 moles of Bh2S7 and 291 moles of FmCO3, you get 3.36 moles of FmS.<br /><br />So, the correct answer is:<br />\[ \boxed{3.36} \]