Question

How many grams of H2O are produced when 35.0 g of NaOH reacts with 17.5 g of CO,?

Answer 1

Answer:

2NaOH + CO2 -> Na2CO3 + H2O

1) Find the moles of each substance

$\eq n(NaOH)=\frac{35.0}{22.99+16.00+1.008\\}\ =\frac{35.0}{39.998} \ = 0.8750437522 moles\\n(CO_{2} ) = \frac{17.5}{12.01+32.00} = \frac{17.5}{44.01} = 0.3976369007 moles$

2) Determine the limitting reagent

$\\NaOH = \frac{0.8750437522}{2} = 0.4375218761$

∴ Carbon dioxide is limitting as it has a smaller value.

3) multiply the limiting reagent by the mole ratio of unknown over known

n(H2O ) = 0.3976369007 × 1/2

             = 0.1988184504 moles

4) Multiply the number of moles by the molar mass of the substance.

m = 0.1988184504 × (1.008 × 2 + 16.00)

   = 0.1988184504 × 18.016

   = 3.581913202 g

Explanation: