Answer:
2.14 g.
Explanation:
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In this case, since the mole ratio between nitrogen and oxygen is 2:5 respectively, we realize we first need the moles of nitrogen given it's atomic mass (14.01 g/mol):
[tex]n_N=0.750g*\frac{1mol}{14.01g}=0.0535molN[/tex]
Now, we calculate the moles of oxygen:
[tex]n_O=0.0535molO*\frac{5molO}{2molN}=0.134molO[/tex]
Then, we compute the moles of oxygen given its atomic mass (16.00 g/mol):
[tex]m_O=0.134molO*\frac{16.00gO}{1molO_2} \\\\m_O=2.14gO[/tex]
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