Answer:
A- 286.3L
Explanation:
Using the ideal gas law equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
According to the information given in this question;
P = 1.14atm
V = ?
n = 15moles
T = 265K
Using PV = nRT
V = nRT ÷ P
V = (15 × 0.0821 × 265) ÷ (1.14)
V = 326.35 ÷ 1.14
V = 286.27 L
V = 286.3L
Chemical equation for above reaction is :
2KClO3 -------> 2KCl + 3O2
Answer:
= 20.82 g of BaCl2
Explanation:
Given,
Volume = 200 mL
Molarity = 0.500 M
Therefore;
Moles = molarity × volume
= 0.2 L × 0.5 M
= 0.1 mole
But; molar mass of BaCl2 is 208.236 g/mole
Therefore; 0.1 mole of BaCl2 will be equivalent to;
= 208.236 g/mol x 0.1 mol
= 20.82 g
Therefore, the mass of BaCl2 in grams required is 20.82 g
