3OsF5+5CoParrow Os3P5+5CoF3 How many grams of CoF3 do I get from 222.2 grams of OsF5 and 111.1 grams of CoP? Os=190F=19Co =59P=31 96 g 101 g 143 g 151 g

To determine the amount of CoF3 produced from 222.2 grams of OsF5 and 111.1 grams of CoP, we need to follow these steps:<br /><br />1. Calculate the molar masses of the reactants and products.<br />2. Determine the limiting reactant.<br />3. Calculate the amount of CoF3 produced based on the limiting reactant.<br /><br />Given:<br />- Os = 190 g/mol<br />- F = 19 g/mol<br />- Co = 59 g/mol<br />- P = 31 g/mol<br /><br />Molar masses:<br />- OsF5 = 190 + 5(19) = 290 g/mol<br />- CoP = 59 + 31 = 90 g/mol<br />- CoF3 = 59 + 3(19) = 98 g/mol<br /><br />Step 1: Calculate the moles of OsF5 and CoP.<br />Moles of OsF5 = 222.2 g / 290 g/mol = 0.765 moles<br />Moles of CoP = 111.1 g / 90 g/mol = 1.234 moles<br /><br />Step 2: Determine the limiting reactant.<br />The balanced chemical equation is:<br />3OsF5 + 5CoP → Os3P5 + 5CoF3<br /><br />From the equation, we can see that 3 moles of OsF5 react with 5 moles of CoP. Therefore, the ratio of OsF5 to CoP is 3:5.<br /><br />To find the limiting reactant, we compare the ratio of the moles of OsF5 to CoP to the ratio of 3:5.<br />Ratio of OsF5 to CoP = 0.765 moles / 1.234 moles = 0.62<br />Since 0.62 is less than 3/5 (0.6), OsF5 is the limiting reactant.<br /><br />Step 3: Calculate the amount of CoF3 produced.<br />From the balanced chemical equation, we can see that 3 moles of OsF5 produce 5 moles of CoF3.<br /><br />Moles of CoF3 produced = (5/3) * 0.765 moles = 1.275 moles<br /><br />Mass of CoF3 produced = 1.275 moles * 98 g/mol = 125.35 g<br /><br />Therefore, the amount of CoF3 produced from 222.2 grams of OsF5 and 111.1 grams of CoP is approximately 125.35 grams.<br /><br />The closest answer choice is 143 g.