Question

Given the following equation, how many grams of PbCO3 will dissolve when exactly 1.0 L of 1.00 M H+ is added to 6.00 g of PbCO3?

When the equation is PbCO3(s) + 2H(+)(aq) = Pb(2+)(aq) + H2O(l) + CO2(g)
Please explain in detail, step by step

Answer 1

To determine how many grams of PbCO3 will dissolve, we first find the moles of PbCO3 in 6.00 g and then confirm that 1.0 L of 1.00 M H+ provides enough H+ ions to react with all of it. All 6.00 g of PbCO3 will dissolve.

Step-by-step explanation:

The student asked how many grams of PbCO3 will dissolve when 1.0 L of 1.00 M H+ is added to 6.00 g of PbCO3, according to the reaction PbCO3(s) + 2H+(aq) = Pb2+(aq) + H2O(l) + CO2(g). To answer this, we need to calculate the number of moles of PbCO3 that correspond to 6.00 g and then determine how many of these moles will react with the provided H+ ions.

First, we calculate the molar mass of PbCO3 (Pb = 207.2, C = 12.01, O = 16.00). The total molar mass is 207.2 + 12.01 + (3 × 16.00) = 267.21 g/mol.

Next, we find the moles of PbCO3 in 6.00 g:
6.00 g ÷ 267.21 g/mol = 0.0225 mol.

Since the stoichiometric ratio between PbCO3 and H+ is 1:2, we need 2 moles of H+ for every mole of PbCO3. Thus, 1.0 L of 1.00 M H+ contains 1.00 mol of H+, which is more than enough to react with 0.0225 mol of PbCO3.

Since all PbCO3 will react, we simply convert the moles back to grams: 0.0225 mol × 267.21 g/mol = 6.00 g. Therefore, all 6.00 g of PbCO3 will dissolve with 1.00 M H+ ions present in 1.0 L of solution.

Answer 2

Answer: 6 grams of lead carbonate wlll get easily dissolve in 1 L of 1.0 M
`H^+` solution.

Step-by-step explanation:

`PbCO_3(s)+2H^+(aq)\rightarrow Pb^2+(aq)+H_2O(l)+CO_2(g)`

Moles of
`H^+` (n):

`1.0 M=\frac{\text{mole of }H^+}{\text{Volume of the solution is L}}=(n)/(1 L)`

n = 1 mole

According to reaction 2 moles of
`H^+` dissolves 1 mole of
`PbCO_3` then , 1 mole will dissolve ;
`(1)/(2)* 1` moles of
`PbCO_3` that is 0.5 moles

Mass of the compund:

=Number of moles of compound × Molar mass of compound

Mass of dissolved
`PbCO_3`:

=
`0.5 mole* 267.21 g/mol=133.605 g`

1 L of 1.0 M
`H^+` solution can dissolve 133.605 grams of lead carbonate. So, 6 grams of lead carbonate wlll get easily dissolve in 1 L of 1.0 M
`H^+` solution.