How many mL of 0.774 M HBr are needed to dissolve 6.73 g of CaCO3? 2HBr(aq) + CaCO3(s) -> CaBr2(aq) + H2O(1) + CO2(g)
_mL

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.

water is used to store them because it would keep the radiation from getting out in the open and causing harm to the workers

the answer would be a