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
Answer:
190.4g
Explanation:
1.6mol of KBr (119.002g KBr/1 mol) = 190.4g
since you want to find grams, take the molar mass of KBr (119.002) per 1 mol and use it as your conversion factor (119.002g KBr/1 mol) which will then cancel out mols and leave you with grams.
Answer:
We need 4.28 grams of sodium formate
Explanation:
<u>Step 1:</u> Data given
MW of sodium formate = 68.01 g/mol
Volume of 0.42 mol/L formic acid = 150 mL = 0.150 L
pH = 3.74
Ka = 0.00018
<u>Step 2:</u> Calculate [base)
3.74 = -log(0.00018) + log [base]/[acid]
0 = log [base]/[acid]
0 = log [base] / 0.42
10^0 = 1 = [base]/0.42 M
[base] = 0.42 M
<u>Step 3:</u> Calculate moles of sodium formate:
Moles sodium formate = molarity * volume
Moles of sodium formate = 0.42 M * 0.150 L = 0.063 moles
<u>Step 4:</u> Calculate mass of sodium formate:
Mass sodium formate = moles sodium formate * Molar mass sodium formate
Mass sodium formate = 0.063 mol * 68.01 g/mol
Mass sodium formate = 4.28 grams
We need 4.28 grams of sodium formate