Answer:
160 g
Explanation:
The formula for the calculation of moles is shown below:
For 
:-
Mass of = 196 g
Molar mass of = 98 g/mol
The formula for the calculation of moles is shown below:
Thus,
According to the given reaction:
1 mole of sulfuric acid reacts with 2 moles of NaOH
So,
2 moles of sulfuric acid reacts with 2*2 moles of NaOH
Moles of NaOH must react = 4 moles
Molar mass of NaOH = 40 g/mol
<u>Mass = Moles*molar mass = 
= 160 g</u>
Answer:
Na₃PO₄ + 3HCl —> 3NaCl + H₃PO₄
The coefficients are: 1, 3, 3, 1
Explanation:
_Na₃PO₄ + __HCl —> __NaCl + _H₃PO₄
The above equation can be balance as follow:
Na₃PO₄ + HCl —> NaCl + H₃PO₄
There are 3 atoms of Na on the left side and 1 atom on the right side. It can be balance by writing 3 before NaCl as shown below:
Na₃PO₄ + HCl —> 3NaCl + H₃PO₄
There are 3 atoms of Cl on the right side and 1 atom on the left side. It can be balance by writing 3 before HCl as shown below:
Na₃PO₄ + 3HCl —> 3NaCl + H₃PO₄
Now, the equation is balanced.
The coefficients are: 1, 3, 3, 1
Answer:
C a B r 2 ( a q ) + N a 2 S O 4 ( a q ) ⟶ 2 N a B r ( a q ) + C a S O 4 ( s )
Explanation:
A precipitation reaction is a type of displacement reaction which a precipitate forms. The precipitate would be in the solid state, different from the other products so it can be separated or removed from the reaction.
C a 2 + ( a q ) + S O 4 2 − ( a q ) ⟶ C a S O 4 ( s )
This is wrong because C a S O 4 is the the only product formed.
C a B r 2 ( a q ) + N a 2 S O 4 ( a q ) ⟶ 2 N a B r ( a q ) + C a S O 4 ( s )
This is the correct option, The precipitate is C a S O 4.
C a 2 + ( a q ) + 2 B r − ( a q ) + 2 N a + ( a q ) + S O 4 2 − ( a q ) ⟶ 2 N a + ( a q ) + 2 B r − ( a q ) + C a S O 4 ( s )
This is the ionic equation for the precipitation reaction
