What is the mass (in grams) of urea (NH2)2CO produced when 7.863 g of NH3 is mixed with 8.8 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

The balanced chemical equation for the synthesis of urea is:

2NH3 + CO2 --> (NH2)2CO + H2O

This tells us that two moles of ammonia (NH3) react with one mole of carbon dioxide (CO2) to produce one mole of urea and one mole of water.

Given: Mass of NH3 = 7.863 g Mass of CO2 = 8.8 g Molar mass of NH3 = 17.03 g/mol Molar mass of CO2 = 44.01 g/mol Molar mass of (NH2)2CO = 60.06 g/mol

Step 1: Calculate the number of moles of each reactant:

Moles of NH3 = mass of NH3 / molar mass of NH3 = 7.863 g / 17.03 g/mol = 0.462 moles Moles of CO2 = mass of CO2 / molar mass of CO2 = 8.8 g / 44.01 g/mol = 0.200 moles

Step 2: Identify the limiting reagent. The balanced equation requires a 2:1 ratio of NH3 to CO2. However, the given mole ratio is 0.462 (NH3) to 0.200 (CO2), which simplifies to approximately 2.31:1. This means there is less CO2 available for the reaction than NH3, relative to what the reaction requires. Therefore, CO2 is the limiting reagent.

Step 3: Calculate the mass of urea produced. According to the balanced equation, one mole of CO2 produces one mole of urea. Therefore, the number of moles of urea produced is equal to the number of moles of the limiting reagent, CO2, which is 0.200 moles.

Mass of (NH2)2CO = moles of (NH2)2CO * molar mass of (NH2)2CO = 0.200 moles * 60.06 g/mol = 12.01 g

So, the mass of urea produced when 7.863 g of NH3 is mixed with 8.8 g of CO2 is 12.01 g.