Answer:
4.37 g of barium sulphate
Explanation:
The reaction equation is;
3BaCl2(aq) + Fe2(SO4)3(aq) ---->3 BaSO4(s) + 2FeCl3(aq)
From the question, the number of moles of both barium chloride and FeSO4 = 125/1000 L × 0.150 M = 0.01875 moles
To find the limiting reactant;
3 moles of barium chloride yields 3 moles of barium sulphate
0.01875 moles of barium chloride yields 3 × 0.01875 moles/3 = 0.01875 moles of barium sulphate
1 mole of iron III sulphate yields 3 moles of barium sulphate
0.01875 molesof iron III sulphate yields 0.01875 moles ×3/1 = 0.05625 moles of barium sulphate
Hence,barium chloride is the limiting reactant
Amount of barium sulphate produced = 0.01875 moles × 233 g/mol = 4.37 g of barium sulphate
Answer:
What are the statements please
Explanation:
Answer:
0.0344 moles and 1.93g.
Explanation:
Molarity is defined as the ratio between moles of a solute (In this case, KOH), and the volume. With molarity and volume we can solve the moles of solute. With moles of solute we can find mass of the solute as follows:
<em>Moles KOH:</em>
15.2mL = 0.0152L * (2.26mol / L) = 0.0344moles
<em>Mass KOH:</em>
0.0344 moles * (56.11g/mol) = 1.93g of KOH
Given:
No of atoms present= 8.022 x 10^23 atoms
Now we know that 1 mole= 6.022 x 10^23 atoms
Hence number of moles present in 8.022 x 10^23 atoms is calculated as below.
Number of moles
= 8.022 x 10^23/6.022x 10^23
=1.3 moles.
Hence we have 1.3 moles present.
