Question

How many molecules of sodium nitrate are produced when 17 g of sodium azide, NaN3, react according to the following equation:

NaN3(aq)+AgNO3(aq) ---> AgN3(s)+NaNO3(aq)

Question options:

3.7x1023


1.6 x 1023


1.85x1023


6.02x1023

Answer 1

The number of molecules of sodium nitrate that reacted is

1.6 x10^23 molecules

calculation

NaN₃(aq) +AgNO₃(aq) → AgN₃(s) + NaNo₃ (aq)

step 1: find moles of NaN₃

moles = mass÷molar mass

from periodic table the molar mass of NaN₃ = 23 +( 3 x 14)= 65 g/mol

moles =17 g÷65 g/mol = 0.262 moles

Step 2: use the mole ratio to calculate the moles of NaNO₃

NaN₃:NaNO₃ is 1:1

Therefore the moles of NaNO₃ is also = 0.262 moles

Step 3: use the Avogadro's law constant to calculate the number of molecules of NaNO₃

That is according to Avogadro's law

1 mole = 6.02 x10^23 molecules

0.262 moles= ? molecules

by cross multiplication

=[(0.262 moles x 6.02 x10^23 molecules) / 1 mole] = 1.6 x10^23 molecules