What is the mass (in grams) of urea (NH2)2CO produced when 7.364 g of NH3 is mixed with 9.819 g of CO2?  Answer to 2 decimal places.2NH3   +   CO2   →   (NH2)2CO   +   H2OMW(NH3) = 17.03 g.mol-1MW(CO2) = 44.01 g.mol-1MW((NH2)2CO) = 60.06 g.mol-1Mw(H2O) = 18.02 g.mol-1

Here's the step-by-step process:

1. Calculate the number of moles of NH3 and CO2 using their respective molar masses:

   Moles of NH3 = 7.364 g / 17.03 g/mol = 0.432 mol Moles of CO2 = 9.819 g / 44.01 g/mol = 0.223 mol

2. From the balanced chemical equation, we can see that the reaction ratio of NH3 to CO2 to (NH2)2CO is 2:1:1. This means that 2 moles of NH3 react with 1 mole of CO2 to produce 1 mole of (NH2)2CO.

3. The limiting reagent is the one that is completely consumed in the reaction. In this case, CO2 is the limiting reagent because we have less moles of CO2 (0.223 mol) than NH3 (0.432 mol).

4. Since the reaction ratio of CO2 to (NH2)2CO is 1:1, the moles of (NH2)2CO produced is equal to the moles of the limiting reagent, which is CO2. So, 0.223 mol of (NH2)2CO is produced.

5. Finally, convert the moles of (NH2)2CO to grams using its molar mass:

   Mass of (NH2)2CO = 0.223 mol * 60.06 g/mol = 13.39 g

So, the mass of urea produced when 7.364 g of NH3 is mixed with 9.819 g of CO2 is 13.39 g.