This is false quantitative is numerical while qualitative is descriptive
Since we assume both reactants have 1 equivalent of H+ and OH- ions, we can balance the moles out. The acid of concentration x will have (x M)(0.035 L) = 0.035x moles of acid. Meanwhile, for the base: (0.432 M)(0.0246 L) = 0.0106 moles of base. Since these must be equivalent:
0.035x = 0.0106x = 0.304 M
Answer:
1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate
Explanation:
Step 1: Data given
Mass of sodium bicarbonate = 2.7 grams
Step 2: The balanced equation
HCl + NaHCO3 ⇔ NaCl + H2O + CO2
Step 3: Calculate moles NaHCO3
moles NaHCO3 =2.7 g / 84 g/mol= 0.032 moles
Step 4: Calculate moles HCl
For 1 mol NaHCO3 we need 1 mol HCl
For 0.032 moles NaHCO3 = 0.032 moles HCl
Step 5: Calculate mass HCl
Mass HCl = moles HCl * molar mass HCl
mass HCl = 0.032 * 36.46 g/mol= 1.17 grams
1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate
Answer- C (water and carbon dioxide)
Answer:
3.14 grams of ammonium thiocyanate must be used to react completely with 6.5 g barium hydroxide octahydrate.
Explanation:
The balance chemical equation is :
Mass of barium hydroxide octahydrate = 6.5 g
Moles of barium hydroxide octahydrate = 
According to reaction, 2 moles of ammonium thiocyanate reacts with1 mole of barium hydroxide octahydrate. The 0.020635 moles of barium hydroxide octahydrate will react with:
Mass of 0.04127 moles of ammonium thiocyanate;
3.14 grams of ammonium thiocyanate must be used to react completely with 6.5 g barium hydroxide octahydrate
