15. determine the percent - composition by ma...
15. determine the percent - composition by mass of zn in zn(oh)₂.\n16. given the equation: si + ga → ga₅si₂\nbalance the equation first. what is the total number of grams of ga₅si₂ that can be formed when 96 grams of si reacts?
Answer
# Explanation:
## Step1: Calculate molar mass of \(Zn(OH)_2\)
The molar mass of \(Zn = 65.38\ g/mol\), \(O=16\ g/mol\), \(H = 1.01\ g/mol\). For \(Zn(OH)_2\), \(M = 65.38+2\times(16 + 1.01)=65.38+2\times17.01=65.38 + 34.02=99.4\ g/mol\).
## Step2: Calculate percent - mass of \(Zn\) in \(Zn(OH)_2\)
The percent - mass of \(Zn\) is \(\frac{65.38}{99.4}\times100\%\).
\[
\begin{align*}
\frac{65.38}{99.4}\times100\%&=\frac{6538}{99.4}\%\\
&\approx65.8\%
\end{align*}
\]
## Step3: Balance the reaction \(Si+Ga\rightarrow Ga_5Si_3\)
The balanced equation is \(3Si + 5Ga\rightarrow Ga_5Si_3\).
## Step4: Calculate moles of \(Si\)
The molar mass of \(Si\) is \(M_{Si}=28.09\ g/mol\). The number of moles of \(Si\), \(n_{Si}=\frac{m}{M}=\frac{96\ g}{28.09\ g/mol}\approx3.42\ mol\).
## Step5: Calculate moles of \(Ga_5Si_3\) formed
From the balanced equation, the mole - ratio of \(Si\) to \(Ga_5Si_3\) is \(3:1\). So the number of moles of \(Ga_5Si_3\) formed, \(n_{Ga_5Si_3}=\frac{3.42\ mol}{3}=1.14\ mol\).
## Step6: Calculate molar mass of \(Ga_5Si_3\)
The molar mass of \(Ga = 69.72\ g/mol\), \(Si = 28.09\ g/mol\). \(M_{Ga_5Si_3}=5\times69.72+3\times28.09=348.6+84.27 = 432.87\ g/mol\).
## Step7: Calculate mass of \(Ga_5Si_3\) formed
\(m_{Ga_5Si_3}=n_{Ga_5Si_3}\times M_{Ga_5Si_3}=1.14\ mol\times432.87\ g/mol\approx493\ g\).
# Answer:
The percent - composition by mass of \(Zn\) in \(Zn(OH)_2\) is approximately \(65.8\%\). The mass of \(Ga_5Si_3\) formed when 96 grams of \(Si\) reacts is approximately \(493\ g\).