25.0 cm3?of 0.050 mol dm-3?sodium carbonate was completely neutralised by 20.0 cm3?of dilute hydrochloric acid.Calculate the concentration in mol dm-3 of the hydrochloric acid.
Answer:
Step 1:?Write the balanced equation for the reaction
Na2CO3??+ ?2HCl ?→? 2NaCl ?+ ?H2O ?+ ?CO2
Step 2:?Determine the moles of the known substance, in this case sodium carbonate. Don't forget to divide the volume by 1000 to convert cm3?to dm3
moles = volume x concentration
amount (Na2CO3) = 0.0250 dm3?x 0.050 mol dm-3?= 0.00125 mol
Step 3:?Use the balanced equation to deduce the mole ratio of sodium carbonate to hydrochloric acid:
1 mol of Na2CO3?reacts with 2 mol of HCl, so the mole ratio is 1 : 2
Therefore 0.00125 moles of Na2CO3?react with 0.00250 moles of HCl
Step 4:?Calculate the concentration of the unknown substance, hydrochloric acid
Calculate the volume of hydrochloric acid of concentration 1.0 mol dm-3 that is required to react completely with 2.5 g of calcium carbonate.
Answer:
Step 1:?Write the balanced equation for the reaction
CaCO3??+ ?2HCl ?→ ?CaCl2??+ ?H2O ?+ ?CO2
Step 2:?Determine the moles of the known substance, calcium carbonate
Step 3:?Use the balanced equation to deduce the mole ratio of calcium carbonate to hydrochloric acid:
1 mol of CaCO3?requires 2 mol of HCl
So 0.025 mol of CaCO3?requires 0.050 mol of HCl
Step 4:?Calculate the volume of HCl required
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