1.44 g of titanium (At. wt. = 48) reacted wi | vedclass

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(B) The reaction is: $1.44 \ Ti + 0.5 \ O_2 \rightarrow Ti_{1.44}O$ The molar mass of $Ti$ is $48 \ g/mol$. The moles of $Ti$ used $= \frac{1.44 \ g}{

Vedclass · Accepted Answer

$1.77$

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(B) The reaction is: $1.44 \ Ti + 0.5 \ O_2 ightarrow Ti_{1.44}O$ The molar mass of $Ti$ is $48 \ g/mol$. The moles of $Ti$ used $= \frac{1.44 \ g}{48 \ g/mol} = 0.03 \ mol$. According to the stoichiometry,$1.44 \ mol$ of $Ti$ produces $1 \ mol$ of $Ti_{1.44}O$. Therefore,$0.03 \ mol$ of $Ti$ produces $\frac{0.03}{1.44} \ mol$ of $Ti_{1.44}O$. The molar mass of $Ti_{1.44}O = (1.44 	imes 48) + 16 = 69.12 + 16 = 85.12 \ g/mol$. The mass $x = 	ext{moles} 	imes 	ext{molar mass} = \frac{0.03}{1.44} 	imes 85.12 = 1.7733 \ g \approx 1.77 \ g$.

$1.44 \ g$ of titanium (At. wt. $= 48$) reacted with excess of $O_2$ to produce $x \ g$ of a non-stoichiometric compound $Ti_{1.44}O$. The value of $x$ is:

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