Answer:
[tex]m_{Si}=37.2gSi[/tex]
Explanation:
Hello,
In this case, for the undergoing balanced chemical reaction:
[tex]SiCl_4 + 2Mg \rightarrow Si + 2MgCl_2[/tex]
We must first identify the limiting reactant given the 225 g of SiCl4 and 101 g of Mg. Thus, we compute the available moles of SiCl4:
[tex]n_{SiCl_4}=225gSiCl_4*\frac{1molSiCl_4}{169.8963gSiCl_4}=1.324molSiCl_4[/tex]
Next, by using the 1:2 mole ratio between SiCl4 and Mg, we compute the moles of SiCl4 consumed by 101 g of Mg:
[tex]n_{SiCl_4}^{consumed}=101gMg*\frac{1molMg}{24.3050gMg} *\frac{1molSiCl_4}{2molMg} =2.08molSiCl_4[/tex]
Thus, since less moles of SiCl4 are available, we can infer it is the limiting reactant whereas the Mg is in excess. In such a way, the produced grams of Si are computed considering the 1:1 molar ratio between SiCl4 and Si:
[tex]m_{Si}=1.324molSiCl_4*\frac{1molSi}{1molSiCl_4} *\frac{28.0855gSi}{1molSi} \\\\m_{Si}=37.2gSi[/tex]
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