Question

1.

5.00 g of butane are combusted in a sealed container containing 2.00 g of oxygen gas. The balanced reaction for the combustion of butane is shown below. How many grams of the excess reagent are left over? (This question first requires you to determine the limiting reagent, and then calculate how much will be leftover). Pease answer in g.

Molar Mass Butane: 58.12 g/mol

Molar Mass Oxygen Gas: 32.0 g/mol

2CH₃CH₂CH₂CH₃ + 13O₂ --> 8CO₂ + 10H₂O

2.

If 10.0 g HCl is reacted with excess NaOH and 12.0 g of NaCl is produced, what is the percent yield of the reaction?

HCl + NaOH --> H₂O + NaCl

Answer 1

sof2017n (1 2 CH3 CH₂ CH₂ C113 + 1302 - 8002 + 70 420 Nt. of Butene som Wt. of oxygen a 2 am m.wt.01 Buime : 58.12.30 / moine m wt of Axygen 32 sm (moc. for limitting Peagent 2 CH₃ CH₂ CH₂ CH₂ t 1302 800₂ + 10 H₂O 2 x 55.12 gm - Bulime *13 (32) om oxygen 8(44) omy + 10 (18) Curson weater droxide 116.34 gm at 416 am 382m + 150 am Here we have som butine crad a sy oxygen . so, for combustion of butane 116.25gm we need 416 can of Butane n o then 5am идJ Mut 4 - Butane 5*476 116.24 . = 17.8g gm Oq. We need. But we have only a om 02 so, O2 is limiting Reagent calculation for overleft butime I excess Butine = 416 gor O2 -> 116.2h am Cahio to the Callio - used Cuttin then a gma 3 Chtio 1 - 5 m - 0.5588 am 2x416 2x 116.24 = 4,64 am (4407 416 0.5588m Cahorused over left Scanned with S CamScanner .

USAHA RA - Null Peet NUOR H₂O + Wt. of Me = 10gm wt-of nace = 12 sm wt. of NaOH2 Excess m.wt of Hel - 36.5 9 imor m.wt of Nace - 58.5 olmos M. wt of Nuor= 40 87moe Helt NuoH - H₂O + nach 1mal Imol 1 mol 1 mol 36.5 gm 40 am is om 58.5 mm If we have, - 36.5em He then 58.5 my nucl e produce soy 10 am rel then & nece 10x58,5 - 36.5 . .. will be = 16.027 am weld produces but in this reaction 12 am Nach produced so In Yield ioos. % Yield .16.627 am - 12 am - 8 12 x 100%. 16.027 - 74.87 % yield cs Scanned with CamScanner