Answer:
44 mL of Na2SO4
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
Ba(NO3)2 (aq) + Na2SO4 (aq) —> BaSO4 (s) + 2NaNO3 (aq)
Step 2:
Determination of the number of mole of Ba(NO3)2 in 20.00 mL of 0.500 M barium nitrate (Ba(NO3)2). This is illustrated below:
Molarity of Ba(NO3)2 = 0.5 M
Volume of solution = 20 mL = 20/1000 = 0.02 L
Mole of solute (Ba(NO3)2) =?
Molarity = mole /Volume
0.5 = Mole of Ba(NO3)2 / 0.02
Cross multiply to express in linear form
Mole of Ba(NO3)2 = 0.5 x 0.02
Mole of Ba(NO3)2 = 0.01 mole
Step 3:
Determination of the number of mole of Na2SO4 that reacted.
Ba(NO3)2 (aq) + Na2SO4 (aq) —> BaSO4 (s) + 2NaNO3 (aq)
From the balanced equation above,
1 mole of Ba(NO3)2 reacted with 1 mole of Na2SO4.
Therefore, 0.01 mole of Ba(NO3)2 will also react with 0.01 mole of Na2SO4.
Step 4:
Determination of the volume of Na2SO4 needed for the reaction. This is illustrated below:
Mole of Na2SO4 = 0.01 mole
Molarity of Na2SO4 = 0.225M
Volume =?
Molarity = mole /Volume
0.225 = 0.01 / volume
Cross multiply to express in linear form
0.225 x Volume = 0.01
Divide both side by 0.225
Volume = 0.01/0.225
Volume of Na2SO4 = 0.044 L
Converting 0.044 L to mL, we have
Volume of Na2SO4 = 0.044 x 1000
Volume of Na2SO4 = 44 mL
Therefore, 44 mL of Na2SO4 is needed for the reaction
m(H2SO4) which is =<span>470g