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Jusaquikie
a chemist runs the balanced chemical reaction shown below by mixing 6.18g o Ca3P2 with 9.49g of H2O, how much of PH3 will be formed asuming 100% yield?
\[Ca _{3}P _{2}(s)+6H _{2}O(l)\rightarrow 3 Ca(OH)_{s}(s)+2 PH _{3}(g)\]
http://openstudy.com/study#/updates/50c27afae4b066f22e104a93 I explain limiting reagent
If there is a 100% yield you would just be determining the theoretical yield, scroll down to the bottom post of that question
i took the grams of Ca3P2/MM and H20 and got the limiting agent as the CaP2 and that is a 1:1 ratio with the 2PH3
once i figure out the ratio is 1:1 then i take the moles of Ca3P2 then multiply by the MM of PH3 to get the grams right?
The ratio between Ca3P2 and PH3 isn't 1:1
i thought i compare the P? P2 in the first to 2P in the second?
for every 1 Ca3P2 you produce 2 PH3
Show me the moles of Ca3P2 and H2O
\[\frac{ 6.18 }{ 182.18 }=.033922\] \[\frac{ 9.49 }{ 18 }=.52722\] \[\frac{ .52722 }{ .033922 }=15.542\] 15>6 so Ca3P2 is the limiting reactant right?
The way I do it is usually convert moles of reagent to product then see which gives me the least product (0.52722/6)2 = (0.033922/1)2 =
doing it this way also gives you the moles of product you have so its not like it is a huge hassle
you produce 2 PH3
ok so i should have (.03392)*2 moles of PH3
then i multiply that by the MM of PH3 and i get the right answer. Thank you very much for your time it's clearer now.