Answer:
173.9 mL of HBr
Explanation:
We'll begin by calculating the number of mole in 6.73 g of CaCO₃. This can be obtained as follow:
Mass of CaCO₃ = 6.73 g
Molar mass of CaCO₃ = 40 + 12 + (16×3)
= 40 + 12 + 48
= 100 g/mol
Mole of CaCO₃ =?
Mole = mass / Molar mass
Mole of CaCO₃ = 6.73 / 100
Mole of CaCO₃ = 0.0673 mole
Next, we shall determine the number of mole of HBr that will react with 6.73 g (i.e 0.0673 mole) of CaCO₃. This can be obtained as follow:
2HBr + CaCO₃ —> CaBr₂ + H₂O + CO₂
From the balanced equation above,
2 moles of HBr reacted with 1 mole of CaCO₃.
Therefore, Xmol of HBr will react with 0.0673 mole of CaCO₃ i.e
Xmol of HBr = 2 × 0.0673
Xmol of HBr = 0.1346 mole
Thus, 0.1346 mole of HBr reacted.
Next, we shall determine the volume of HBr needed for the reaction. This can be obtained as follow:
Mole of HBr = 0.1346 mole
Molarity of HBr = 0.774 M
Volume =?
Molarity = mole / Volume
0.774 = 0.1346 / volume
Cross multiply
0.774 × volume = 0.1346
Divide both side by 0.774
Volume = 0.1346 / 0.774
Volume = 0.1739 L
Finally, we shall convert 0.1739 L to mL. This can be obtained as follow:
1 L = 1000 mL
Therefore,
0.1739 L = 0.1739 L × 1000 mL / 1 L
0.1739 L = 173.9 mL
Thus, 173.9 mL of HBr is needed for the reaction.
