2020-09-20T09:49:20-04:00
Given this unbalanced (Check) reaction, how many moles of Chromium (III) Sulfide will you have if you react 10.0 grams of Chromium (III) Oxide with excess Hydrogen Sulfide? Follow Significant Figures
1
2020-09-21T06:47:48-0400
n ( C r 2 O 3 ) = n ( C r 2 S 3 ) n(Cr_2O_3) = n(Cr_2S_3) n ( C r 2 O 3 ) = n ( C r 2 S 3 )
n ( C r 2 O 3 ) = m ( C r 2 O 3 ) M ( C r 2 O 3 ) n(Cr_2O_3) = {m(Cr_2O_3) \over M(Cr_2O_3)} n ( C r 2 O 3 ) = M ( C r 2 O 3 ) m ( C r 2 O 3 )
n ( C r 2 O 3 ) = 10.0 g 151.99 g / m o l = 0.066 m o l n(Cr_2O_3) = {10.0g \over 151.99g/mol} = 0.066 mol n ( C r 2 O 3 ) = 151.99 g / m o l 10.0 g = 0.066 m o l
m ( C r 2 S 3 ) = n ( C r 2 S 3 ) ∗ M ( C r 2 S 3 ) m(Cr_2S_3) = n(Cr_2S_3)*M(Cr_2S_3) m ( C r 2 S 3 ) = n ( C r 2 S 3 ) ∗ M ( C r 2 S 3 )
m ( C r 2 S 3 ) = 0.066 m o l ∗ 200.19 g / m o l = 13.2 g m(Cr_2S_3) = 0.066mol*200.19g/mol = 13.2 g m ( C r 2 S 3 ) = 0.066 m o l ∗ 200.19 g / m o l = 13.2 g Answer: 13.2 g
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