Answer:
[tex]m_{SO_2}=0.175gSO_2[/tex]
Explanation:
Hello,
In this case, since the first step is to write the properly balanced chemical reaction:
[tex]P_4S_3 (s) + 8O_2 (g)\rightarrow P_4O_{10}(s) + 3SO_2 (g)[/tex]
We can see that given the 0.200 g of P4S3 (molar mass 220 g/mol) the mole ratio between it and SO2 (molar mass 64 g/mol) is 1:3, therefore, the produced mass of SO2 turns out:
[tex]m_{SO_2}=0.200gP_4S_3*\frac{1molP_4S_3}{220gP_4S_3} *\frac{3molSO_2}{1molP_4S_3} *\frac{64gSO_2}{1molSO_2} \\\\m_{SO_2}=0.175gSO_2[/tex]
Best regards.
