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Fe<sub>2</sub>O<sub>3</sub>+3CaS = Fe<sub>2</sub>S<sub>3</sub>+3CaO How many grams of Fe<sub>2</sub>O<sub>3</sub> are needed to produce 51.9 grams of CaO?
Solution:Fe2O3 + 3CaS = Fe2S3 + 3CaOx g 51,9 g(Fe2O3) = 56*2+16*3= 160g/mol(CaO)=40+16=56g/molXmol(CaO)= 51,9/56= 0,96Xmol(Fe2O3)=0,92/3=0,308Xg(Fe2O3)=160*0,308/1=49,28Answer:m(Fe2O3)=49,28g
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What is the ΔH° for the following reaction (in kJ/mol)? 2 Na <sub>(s) </sub>+ 2 H<sub>2</sub>O<sub> (l)</sub> →<sub> </sub> 2 NaOH<sub> (aq) </sub>+ H<sub>2</sub> <sub>(g)</sub>
From the given info, we know the amount of heat released when 2 moles of Na is reacted. We're asked for the heat per 0.5 moles. So, (368.6 kJ/2mol)0.5000 mol = 92.15...
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show that compounds N<sub>2</sub>O, NO, N<sub>2</sub>O<sub>3</sub>,NO<sub>2</sub> and N<sub>2</sub>O<sub>5</sub> illustrate the law of multiple proportions?
The law of multiple proportion displays that when 2 elements combine with each other forming 2 or more chemical compound then different amount of 1 element which react with definite...
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N<sub>2(g)</sub>+3H<sub>2(g)</sub>->2NH<sub>3(g)</sub> Given:H<sub>2(g)</sub>=131G N<sub>2(g)</sub>=192 NH<sub>3(g)</sub>0=193. S<sup>0 </sup>values in J/K.mol
Value =199 J/K.mol
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How many mL of 0.250 M KOH will react with 15.0 mL of 0.350 M H<sub>2</sub>SO<sub>4</sub>? The reaction is : 2KOH + H<sub>2</sub>SO<sub>4</sub> → K<sub>2</sub>SO<sub>4</sub> + 2H<sub>2</sub>O
mL of 0.250 M KOH=5.5
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In the following reaction, how many liters of O<sub>2</sub> will produce 43.62 liters of CO<sub>2</sub> at STP? C<sub>3</sub>H<sub>8</sub> + 5 O<sub>2</sub> 3 CO<sub>2</sub> + 4 H<sub>2</sub>O
Litres of O2=72.7L O2
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How many moles of CO<sub>2</sub> will be produced when you burn 12.5 moles of C<sub>2</sub>H<sub>4</sub>? ( C<sub>2</sub>H<sub>4</sub> + 3 O<sub>2</sub> ---> 2 CO<sub>2</sub> + 2 H<sub>2</sub>O )
Mole ratio C2H4:CO2=1:2= 12.5×2=25moles of CO2
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Can you please help me balance this equation?Na[sub]3[/sub]PO[sub]4[/sub] + Ba(NO[sub]3[/sub])[sub]2[/sub] = Ba[sub]3[/sub](PO[sub]4[/sub])[sub]2[/sub] + NaNO[sub]3[/sub]
The correct balanced equation is: Na3PO4 + Ba(NO3)2 = BaPO4 + Na3(NO3)2
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2HCl + Ba(OH) <sub>2</sub> --> BaCl <sub>2</sub> + 2H <sub>2</sub>O If you have 80.0 grams of HCl and 50.0 grams of Ba(OH) <sub>2</sub>, how many grams of BaCl <sub>2</sub> can you make?
2HCL = 2molBa(OH)2 = 1molBaCl2 = 1 mol2H2O = 2molMolar Mass BaCL2 = 208.23g/molMolar Mass of Ba(OH)2 = 171.34g/molUsing the ratio(208.3g/mol Bacl2/ 171.34Ba(OH)2) ×50 = 60.78557= 60.79 Bacl2Molar Mass of...
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CH <sub>3 </sub>C(CH <sub>3 </sub>) <sub>2 </sub>CH <sub>2 </sub>CH=CHCH <sub>2 </sub>CH <sub>3 </sub>what is the correct IUPAC name
The correct IUPAC name is 6,6-dimethylhept-3-ene.
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