Question

1: Calcule a quantidade de H2SO4 usada para reagir com 300g de CA (OH)2
H2So4 + Ca(OH)2 CASO4 + 2H2O

2: Calcule a quantidade de CASO4 formada a partir de 200g de CA(OH)2 com 60% de rendimento.
H2SO4 + Ca(OH)2 CASO4 + 2H20

3: Calcule a quantidade de CA(OH)2 usada para reagir com 500g de H2SO4 90% puro.
H2SO4 + CA(OH)2 CASO4 + 2H20

4: Calcule a quantidade de CA3(PO4)2 formada a partir de 300g de H3PO4 com 60% de rendimento
CA(OH) + H3PO4 CA3(PO4)2 + H20

5: Calcule a quantidade de H2SO4 usada para reagir com 500g de FE(OH)3 com 90% de pureza
H2SO4 + FE(OH)3 FE2(SO4)3 + H20

edit: Preciso fazer essas atividades ate segunda feira e n entendi. Se souber pff responda e com explicação para eu entender.

Answer 1

$\large\boxed{\begin{array}{l}\rm Para\,resolver\,estas\,quest\tilde oes\\\rm usarei\,a\,tabela\,peri\acute odica\,do\,vestibular\\\rm da\,UERJ.\\\rm A\,qual\,arrendonda\,os\,valores\\\rm das\,massas\,at\hat omicas\\\rm de\,alguns\,elementos.\end{array}}$

$\large\boxed{\begin{array}{l}\rm 1)\rm H_2SO_4+Ca(OH)_2\longrightarrow CaSO_4+2H_2O\\\rm 98\,g\,de\,H_2SO_4\longrightarrow 74\,g\,de\,Ca(OH)_2\\\rm x\,g\,de\,H_2SO_4\longrightarrow 300\,g\,de\,Ca(OH)_2\\\rm 74x=29400\\\rm x=\dfrac{29400}{74}\\\\\rm x=397,29\,g\,de\,H_2SO_4\end{array}}$

$\boxed{\begin{array}{l}\rm 2)~ H_2SO_4+Ca(OH)_2\longrightarrow CaSO_4+2H_2O\\\rm74\,g\,de\,Ca(OH)_2\longrightarrow136\,g\,de\,CaSO_4\\\rm 200\,g\,de\,Ca(OH)_2\longrightarrow y\,g\,de\,CaSO_4\\\rm 74y=200\cdot136\\\rm 74y=27200\\\rm y=\dfrac{27200}{74}\\\\\rm y=367,56\,g\,de\, CaSO_4\\\underline{\rm C\acute alculo\,com\,60\%\,de\,rendimento\!:}\\\rm 367,56\longrightarrow 100\%\\\rm x\longrightarrow 60\%\\\rm 100x=367,56\cdot60\\\rm 100x=220053,6\\\rm x=\dfrac{22053,6}{100}\\\\\rm y=220,536\end{array}}$

$\large\boxed{\begin{array}{l}\rm 3)~H_2SO_4+Ca(OH)_2\longrightarrow CaSO_4+2H_2O\\\rm 90\%\,de\,500=0,9\cdot500=450\\\rm98\,g\,de\,H_2SO_4\longrightarrow74\,g\,de\, Ca(OH)_2\\\rm450\,g\,de\,H_2SO_4\longrightarrow x\\\rm 98x=33300\\\rm x=\dfrac{33300}{98}\\\\\rm x=339,79\,g\,de\,Ca(OH)_2\end{array}}$

$\large\boxed{\begin{array}{l}\rm 4)\\\rm 3Ca(OH)+2H_3PO_4\longrightarrow Ca_3(PO_4)_2+6H_2O\\\rm196\,g\,de\,H_3PO_4\longrightarrow310\,g\,de\,Ca_3(PO_4)_2\\\rm 300\,g\,de\,H_3PO_4\longrightarrow z\\\rm 196z=93000\\\rm z=\dfrac{93000}{196}\\\\\rm z=474,48\,g\,de\,Ca_3(PO_4)_2\\\rm 474,48\longrightarrow 100\%\\\rm x\longrightarrow 60\%\\\rm x=\dfrac{60\cdot474,48}{100}\\\\\rm x=284,688\,g\,de\,Ca_3(PO_4)_2\end{array}}$

$\large\boxed{\begin{array}{l}\rm 5)\\\underline{\rm Equac_{\!\!\!,}\tilde ao\,balanceada}\\\rm 3H_2SO_4+2Fe(OH)_3\longrightarrow Fe_2(SO_4)_3+6H_2O\\\rm 90\%\,de\,500=0,9\cdot500=450\\\rm 294\,g\,de\,H_2SO_4\longrightarrow214\,g\,de\,Fe(OH)_3\\\rm x\longrightarrow 450\,g\,de\,Fe(OH)_3\\\rm 214x=132300\\\rm x=\dfrac{132300}{214}\\\\\rm x=618,22\,g\,de\,H_2SO_4\end{array}}$