# Chapter 4 - Exercises - Page 188: 52

Limiting reactant: $SiO_2$ Theoretical yield: 72.84 kg Percent yield: 90.7%

#### Work Step by Step

- Calculate or find the molar mass for $SiO_2$: $SiO_2$ : ( 28.09 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 60.09 kg/kmol - Using the molar mass as a conversion factor, find the amount in kmol: $$155.8 \space kg \times \frac{1 \space kmol}{ 60.09 \space kg} = 2.593 \space kmol$$ - Calculate or find the molar mass for $C$: $C$ : 12.01 kg/kmol - Using the molar mass as a conversion factor, find the amount in kmol: $$78.3 \space kg \times \frac{1 \space kmol}{ 12.01 \space kg} = 6.52 \space kmol$$ Find the amount of product if each reactant is completely consumed. $$2.593 \space kmol \space SiO_2 \times \frac{ 1 \space kmol \ Si }{ 1 \space kmol \space SiO_2 } = 2.593 \space kmol \space Si$$ $$6.52 \space kmol \space C \times \frac{ 1 \space kmol \ Si }{ 1 \space kmol \space C } = 6.52 \space kmol \space Si$$ Since the reaction of $SiO_2$ produces less $Si$ for these quantities, it is the limiting reactant. - Calculate or find the molar mass for $Si$: $Si$ : 28.09 kg/kmol - Using the molar mass as a conversion factor, find the mass in kg: $$2.593 \space kmol \times \frac{ 28.09 \space kg}{1 \space kmol} = 72.84 \space kg$$ $$Percent \space yield = \frac{ 66.1 }{ 72.84 } \times 100\% = 90.7 \%$$

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