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:1. Calculate the molar masses of the reactants and products.2. Determine the limiting reactant.3. Calculate the amount of CoF3 produced based on the limiting reactant.Given:- Os = 190 g/mol- F = 19 g/mol- Co = 59 g/mol- P = 31 g/molMolar masses:- OsF5 = 190 + 5(19) = 290 g/mol- CoP = 59 + 31 = 90 g/mol- CoF3 = 59 + 3(19) = 98 g/molStep 1: Calculate the moles of OsF5 and CoP.Moles of OsF5 = 222.2 g / 290 g/mol = 0.765 molesMoles of CoP = 111.1 g / 90 g/mol = 1.234 molesStep 2: Determine the limiting reactant.The balanced chemical equation is:3OsF5 + 5CoP → Os3P5 + 5CoF3From 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.To find the limiting reactant, we compare the ratio of the moles of OsF5 to CoP to the ratio of 3:5.Ratio of OsF5 to CoP = 0.765 moles / 1.234 moles = 0.62Since 0.62 is less than 3/5 (0.6), OsF5 is the limiting reactant.Step 3: Calculate the amount of CoF3 produced.From the balanced chemical equation, we can see that 3 moles of OsF5 produce 5 moles of CoF3.Moles of CoF3 produced = (5/3) * 0.765 moles = 1.275 molesMass of CoF3 produced = 1.275 moles * 98 g/mol = 125.35 gTherefore, the amount of CoF3 produced from 222.2 grams of OsF5 and 111.1 grams of CoP is approximately 125.35 grams.The closest answer choice is 143 g.