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Sagot :
Answer: 4.1 g of barium precipitated.
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar Mass}}[/tex]
Given : moles of barium = 0.030
Molar mass of barium = 137 g/mol
[tex]0.030=\frac{x}{137}[/tex]
x= 4.1 g
Thus there are 4.1 g of barium that precipitated.
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