0.25 g of sodium hydroxide reacts with hydrogen bromide to form sodium bromide. How many grams of sodium bromide can be produced?
NaOH (g) +HBr→ NaBr (g) + H2O
(Molar masss: Na = 23 g/mol; O = 16 g/mol; H= 1 g/mol; Br= 80 g/mol)
First calculate the moles of NaOH:
moles = mass / molar mass
moles = 0.25 g / 40 g/mol (NaOH molar mass)
moles = 0.00625 mol
From the balanced equation, we can see that the ratio of moles of NaOH and NaBr is 1:1. This means that 0.00625 mol of NaBr will be produced.
Now, calculate the mass of NaBr:
mass = moles x molar mass
mass = 0.00625 mol x 102 g/mol (NaBr molar mass)
mass = 0.6375 g
Therefore, 0.25 g of NaOH can produce 0.6375 g of NaBr.
To calculate the number of grams of sodium bromide that can be produced from 0.25 g of sodium hydroxide, we need to use the molar ratios of the reactants and products.
1. Calculate the molar mass of sodium hydroxide (NaOH):
Na = 23 g/mol
O = 16 g/mol
H = 1 g/mol
Molar mass of NaOH = (23 g/mol) + (1 g/mol) + (16 g/mol) = 40 g/mol
2. Convert the given mass of sodium hydroxide to moles:
Moles of NaOH = Mass of NaOH / Molar mass of NaOH
= 0.25 g / 40 g/mol
= 0.00625 mol
3. Use the balanced chemical equation to determine the molar ratio between sodium hydroxide and sodium bromide:
NaOH (g) + HBr → NaBr (g) + H2O
From the equation, we can see that the molar ratio between NaOH and NaBr is 1:1.
4. Since the molar ratio between NaOH and NaBr is 1:1, the number of moles of NaBr produced will be the same:
Moles of NaBr = 0.00625 mol
5. Calculate the mass of sodium bromide produced:
Mass of NaBr = Moles of NaBr × Molar mass of NaBr
= 0.00625 mol × 80 g/mol
= 0.5 g
Therefore, 0.25 g of sodium hydroxide will produce 0.5 g of sodium bromide.