What mass of sodium chloride (NaCl) forms when 7.5 g of sodium carbonate (Na2CO3) reacts with a dilute solution of hydrochloric acid (HCl) ?

What mass of sodium chloride (NaCl) forms when 7.5 g of sodium carbonate (Na2CO3) reacts with a dilute solution of hydrochloric acid (HCl) ?

Next, cancel like terms and units. Which of the following shows the correct terms and units canceled?

First Option

The mass of NaCl formed is 8.307 grams

calculation

step 1: write the equation for reaction

Na₂CO₃ + 2HCl → 2 NaCl + CO₂ + H₂O

Step 2: find the moles of Na₂CO₃

moles = mass/molar mass

The molar mass of Na₂CO₃ is = (23 x2) + 12 + (16 x3) = 106 g/mol

moles = 7.5 g/106 g/mol = 0.071 moles

Step 3: use the mole ratio to determine the mole of NaCl

Na₂CO₃:NaCl is 1:2 therefore the moles of NaCl = 0.07 x2 = 0.142 moles

Step 4: calculate mass of NaCl

mass = moles x molar mass

the molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

mass = 0.142 moles x 58.5 g/mol = 8.307 grams