The question has missing information. At part 1 it is "Write a balanced chemical equation, including physical state symbols, for the decomposition of solid calcium carbonate (CaCO3) into solid calcium oxide and gaseous carbon dioxide."
Part 2. "Suppose 19.0 L of carbon dioxide gas are produced by this reaction, at a temperature of 290.0°C and pressure of exactly 1 atm. Calculate the mass of calcium carbonate that must have reacted (...)"
Answer:
41.0 g
Explanation:
1. Calcium oxide has molecular formula CaO and carbon dioxide CO₂, thus, the reaction will be:
CaCO₃(s) → CaO(s) + CO₂(g)
The equation is already balanced because there's the same number of each element on both sides.
2. First, let's calculate the number of moles of CO₂ produced by the ideal gas law:
PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.082 atm.L/mol.K), and T is the temperature (290°C = 273 = 563 K).
1*19 = n*0.082*563
46.166n = 19
n = 0.4116 mol
By the stoichiometry of the reaction:
1 mol of CaCO₃ ------ 1 mol of CO₂
x ----- 0.4116 mol
By a simple direct three rule:
x = 0.4116 mol of CaCO₃.
The molar mass of the calcium carbonate is 100 g/mol, thus the mass (m) is the number of moles multiplied by it:
m = 0.4116*100
m = 41.16 g = 41.0 g
. For He, The outermost electrons are 2.
( where, n = 2, 3, 4, 5, 6 and 7 respectively). For all the other gases, the outermost electrons are 8.