A is a solution of HNo3 of unknown concentration while is a solution of sodium hydroxide containing 4.00 g/dmcube of solution 5.00cm cube of B using methyl orange as an indicator.from the result and information provide above write a balanced equation for the reaction.
To write a balanced equation for the reaction, we first need to determine the reaction between HNO3 (solution A) and sodium hydroxide (solution B).
Given that methyl orange is used as an indicator, we can assume that a neutralization reaction has occurred, leading to the formation of water (H2O) and a salt.
From the information provided, we know the volume of solution B (5.00 cm³) and the concentration of sodium hydroxide within it (4.00 g/dm³).
To find the number of moles of sodium hydroxide in solution B, we need to convert the mass to moles using its molar mass. The molar mass of NaOH is approximately 40.00 g/mol (22.99 g/mol for Na + 16.00 g/mol for O + 1.01 g/mol for H).
Number of moles of NaOH = mass / molar mass
= (4.00 g) / (40.00 g/mol)
= 0.100 mol
Next, we can determine the number of moles of HNO3 in solution A by balancing the equation. We can use the mole ratio between NaOH and HNO3 from the balanced equation to find the number of moles of HNO3.
The balanced equation for the reaction between NaOH and HNO3 is:
HNO3 + NaOH → NaNO3 + H2O
From the equation, we can see that 1 mole of HNO3 reacts with 1 mole of NaOH to produce 1 mole of NaNO3 and 1 mole of H2O.
Therefore, the number of moles of HNO3 in solution A is also 0.100 mol.
So, the balanced equation for the reaction is:
HNO3 + NaOH → NaNO3 + H2O