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following balanced chemical
reaction:
Ca2+
(aq)+ CO3
2--
(aq) ⇄CaCO3(S)
The amount of dissolved CaCO3 can be
increased on adding: CaCO3(S)
KNO3(S)
Na2CO3(S)
CH3COOH(S)
According to Le Chatelier's principle, When any system at equilibrium is subjected to change in concentration, temperature, volume or pressure , then the system readjusts itself to counteract (partially) the...
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What is the percent yield of the following reaction if 60 grams of CaCO3 is heated to give 15 grams of CaO? CaCO3→CaO + CO2
Percent Yield =45%
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1.60 g O2 is reacted in excess S , if the percent yield of the reaction is 95.0% what is the yield of SO2
The reaction between O2 and S can be shown as following:S + O2 = SO2The theoretical yield of SO2 produced in the reaction can be calculated using the equation:m(SO2) =...
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What volume of 0.1248 M HCl should be added to a 5.0 g sample of CaCO3 for the CaCO3 to react completely.
CaCO3 (s) + 2 HCl ==>> CaCL2 + H2O + CO2 (g)
As one can see from the reaction equation, two moles of HCl reacts with one mole of calcium carbonate: n(CaCO_3 )=n(HCl)/2. The number of the moles of HCl equals: n(HCl)=c(HCl)·V(HCl)=0.1248...
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combining 0.385 mol of feo3 with excess carbon produced 16.5g of fe.
what is the actual yield of iron in moles?
what was the theoretical yield of iron in moles?
what was the percent yield?
.n (FeO3)=0.385 mol .m (Fe)=16.5 g .n(Fe)theor=? %=? FeO3+ 3C=Fe+3CO .n (Fe)theor= n (FeO3)=0.385 mol Amount of substance of iron n (Fe)=m/Ar= 16.5/56=0.295 mol %= n(Fe)/ n (Fe) theor= 0.295/0.385...
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How many grams of silver bromide, AgBr, are produced if 15.0 mL of 0.350 M CaBr2(aq) completely react given the reaction:
2 AgNO3(aq) + CaBr2(aq) → Ca(NO3)2(aq) + 2 AgBr(s)
n(CaBr2)= V(CaBr2)×c(CaBr2)=0.015×0.35=0.00525 moln(AgBr)=2×n(CaBr2)=2×0.00525=0.0105 molm(AgBr)=n(AgBr)×M(AgBr)=0.0105×188=1.974 g.
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A 43.2 sample of CaCO3 was treated with an excess of aqueous H2SO4, producing calcium sulfate (CaSO4), 3.65 g of water and CO2(g). What was the % yield of H2O?
According to the reaction:CaCO3 + H2SO4 = CaSO4 + H2O + CO2From here, the theoretical yield of water equals::mt (H2O) = m(CaCO3) × Mr(H2O) / Mr(CaCO3) = 43.2 g ×...
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C2H4 + O2 > CO2 + H2O
A. What is my theoretical yield of H2O if I start with 9.35 g of O2?
B. If I actually produced 2.63 g of H2O, what is my percent yield?
9.35g XgC2H4 + 3O2 > 2CO2 + 2H2O 3*32g 2*18gA: m(theor H2O) = X = 9.35*2*18/(3*32) = 3.506g.B: η(H2O) = m(theor H2O)/m(pract H2O) = 2.63g/3.506g*100% = 75.01% = 75%.
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Quicklime, CaO, is produced by the thermal decomposition of calcium carbonate, CaCO3. Calculate the volume of CO2 at STP produced from the decomposition of 152.0g of CaCO3.
34.02 l
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How many grams of AgBr will form when 45.0 mL of 0.842 M CaBr2 react?
CaBr2(aq) + 2 AgNO3(aq) → 2 AgBr(s) + Ca(NO3)2(aq)
Solution:Molar mass of AgBr = 187.77 g/molmass of AgBr = M of CaBr2× V of CaBr2 / 1000ml × n of AgBr × molar mass of AgBr = 0.842molCaBr2× 45mlCaBr2/1000ml×2×187.77g/molAgBr...
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