The number of grams of oxygen in 32.2 g of N | vedclass

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Question

(B) The atomic masses of elements are: $Na = 23 \ g/mol$, $S = 32 \ g/mol$, $O = 16 \ g/mol$, $H = 1 \ g/mol$. The molecular mass of $Na_2SO_4 \cdot 1

Vedclass · Accepted Answer

$22.4$

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(B) The atomic masses of elements are: $Na = 23 \ g/mol$, $S = 32 \ g/mol$, $O = 16 \ g/mol$, $H = 1 \ g/mol$. The molecular mass of $Na_2SO_4 \cdot 10H_2O = (2 	imes 23) + 32 + (4 	imes 16) + (10 	imes 18) = 46 + 32 + 64 + 180 = 322 \ g/mol$. Given mass of $Na_2SO_4 \cdot 10H_2O = 32.2 \ g$. Moles of $Na_2SO_4 \cdot 10H_2O = \frac{32.2 \ g}{322 \ g/mol} = 0.1 \ mol$. In one mole of $Na_2SO_4 \cdot 10H_2O$, there are $4$ oxygen atoms in $Na_2SO_4$ and $10$ oxygen atoms in $10H_2O$, totaling $14$ oxygen atoms. Moles of oxygen atoms $= 0.1 \ mol 	imes 14 = 1.4 \ mol$. Mass of oxygen $= 1.4 \ mol 	imes 16 \ g/mol = 22.4 \ g$.

The number of grams of oxygen in $32.2 \ g$ of $Na_2SO_4 \cdot 10H_2O$ is approximately (in $g$)

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