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.

it would be d. henry moseley

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