Question

21g of MgCO3 is dissolved in 2.5 L of 0.1 M HCl solution. the volume (in mL) of co2 liberated at 1 atm and 300K

Answer 1

Answer:

sorry I don't know

forget me

Answer 2

Answer:

The volume of the $CO_{2}$ ( carbon dioxide ) liberated ( in ml ) calculated is $6150ml$.

Explanation:

Given data,

The mass of $MgCO_{3}$, m = $21g$

The volume of the $HCl$ solution, V = $2.5L$

The molarity of the $HCl$ solution = $0.1M$

The pressure, P = $1atm$

The temperature, T = $300K$

The carbon dioxide's ( $CO_{2}$ ) volume ( in ml ) liberated =?

The reaction of $MgCO_{3}$ dissolved in $HCl$ follows:

• $MgCO_{3} +2 HCl \rightarrow MgCl_{2} + H_{2}O + CO_{2}$

As we can see in the reaction, one mole of $MgCO_{3}$ liberates one mole of $CO_{2}$.

Therefore, we can say that;

• The moles of $MgCO_{3}$ = moles of $CO_{2}$.

As we know,

• The molar mass of $MgCO_{3}$ = $84g/mol$

Therefore,

• The number of moles of $MgCO_{3}$, n = $\frac{given mass}{molar mass}$ = $\frac{21}{84}$ = $0.25mol$

Also,

• The number of moles of $CO_{2}$, n = $0.25mol$

Now, according to the ideal gas equation:

• $PV = nRT$

Here, R = gas constant = $0.082L-atm/K-mol$

Thus,

• V = $\frac{nRT}{P}$
• V = $\frac{0.25\times 0.082\times 300}{1}$
• V = $6.15L$

Convert L into ml

• $1 L$ = $1000ml$
• $6.15L$ = $(6.15\times 1000)ml$ = $6150ml$

Hence, the volume of the $CO_{2}$ liberated ( in ml ) = $6150ml$.