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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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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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How many moles of NO<sub>2</sub> would result from mixing 1.5 mole of N<sub>2</sub> with 2.5 moles of O<sub>2</sub>? N<sub>2</sub> + 2 O<sub>2</sub> ----> 2 NO<sub>2</sub>
1.5 moles of N2 and 2.5 moles of O2 Then O2 is limiting Reagent .2 moles O2 gives 2 moles of NO2 .2.5 moles of O2 gives 2.5 moles of...
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If you have 9.7 moles of O<sub>2</sub> how many moles of water could you make? C<sub>3</sub>H<sub>8</sub>+ 5O<sub>2</sub> → 3CO<sub>2</sub> + 4H<sub>2</sub>O
Solution:Balanced chemical equation:C3H8 + 5O2 → 3CO2 + 4H2OAccording to the equation above: n(O2)/5 = n(H2O)/4Thus,n(H2O) = 4 × n(O2) / 5 = (4 × 9.7 mol) / 5 = 7.76 moln(H2O) =...
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(NH<sub>4</sub>)<sub>2</sub>CO<sub>3</sub> + H<sub>2</sub>SO<sub>4</sub> → (NH<sub>4</sub>)<sub>2</sub>SO<sub>4</sub> + CO<sub>2</sub> + H<sub>2</sub>O net ionic equation
2 NH4 + + CO3 -2+ 2H + +SO4 2-= 2 NH4 + +SO4 2-+ CO2+H2O2NH4++CO3−2+2H++SO42−=2NH4++SO42−+CO2+H2O
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