Question

What mass of Si, in grams, can be produced from the reaction below starting with 225 g SiCl4 and 101 g Mg? SiCl4 + Mg  Si + MgCl2 Given: 1 mol SiCl4 = 169.8963 g SiCl4 1 mol Mg = 24.3050 g Mg 1 mol Si = 28.0855 g Si

Answer 1

`m_(Si)=37.2gSi`

Step-by-step explanation:

Hello,

In this case, for the undergoing balanced chemical reaction:

`SiCl_4 + 2Mg \rightarrow Si + 2MgCl_2`

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:

`n_(SiCl_4)=225gSiCl_4*(1molSiCl_4)/(169.8963gSiCl_4)=1.324molSiCl_4`

Next, by using the 1:2 mole ratio between SiCl4 and Mg, we compute the moles of SiCl4 consumed by 101 g of Mg:

`n_(SiCl_4)^(consumed)=101gMg*(1molMg)/(24.3050gMg) *(1molSiCl_4)/(2molMg) =2.08molSiCl_4`

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:

`m_(Si)=1.324molSiCl_4*(1molSi)/(1molSiCl_4) *(28.0855gSi)/(1molSi) \\\\m_(Si)=37.2gSi`

Best regards.