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Based on the reaction below. How many grams of water would be form from 0.33 moles of C4H10 reacting with an unlimited amount of oxygen gas? (3pts) C4H10 + 6O2 --> 4CO2 + 5H2O
Mole ratio C4H10:H2O= 1:5= 0.33×5= 1.65molesMolar mass of H2O= 18.02= 18.02×1.65=29.733g
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How many grams of water are formed from 20 grams of C5H12 by using this equation C5H12+8O and 5CO2+6H2O
C5H12+8O = 5CO2+6H2O n = m/M m = M·n n (H2O) = 6·n (C5H12) M (C5H12) = 72 g/mol M (H2O) = 18 g/mol n (C5H12) = 20/72 = 0.28...
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What is the concentration (in mol / dm3) of KCl in the solution formed by mixing 25 mL of KCl solution, 0.100 M concentration with 50 mL of KCl solution, 0.100 M concentration?
The concentration of both solutions is equal. Therefore, we will get 75 ml of 0.100 M KCl solution.0.100 M = 0.100 mol/dm3.
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What is the effect of a small amount of impurity on the melting point of an organic compound?
Decreases melting point
Elevates melting point
Stagnates melting point
None of the above
A small amount of impurity decreases melting point an organic compound
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When heated, KClO3 decomposes into KCl and O2.
2 KClO3 ⟶ 2 KCl + 3 O2
If this reaction produced 53.7 g KCl, how many grams of O2 were produced?
The molar mass of potassium chloride is 39+35.5 = 74.5. Thus formed 53.7/74.5 = 0.72 mol oxygen. Based on the oxygen reaction equation formed (0.72*2)/3 = 0.48 mol or 0.48*22.4...
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C5H12 + 8O2 > 5CO2 + 6H2O
How many grams of carbon dioxide will form from the combustion 320.0 grams of C5H12 with excess oxygen?
5n(C5H12)=n(CO2)m(CO2)=44,00 g/mole * n(CO2)m(C5H12)=72,00 g/mole * n(C5H12)n(C5H12)=4,444 molesn(CO2)=22,22 molesm(CO2)=977,68 g
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Calculate the mass (in grams) of H2O produced from the reaction of 56.20 g C4H10 with 85.30 g O2.
C4H10 (g) + O2 (g) → CO2 (g) + H2O (l)
C4H10(g) + 6,5O2(g) → 4CO2(g) + 5H2O(l)n(C4H10)=m(C4H10)/M(C4H10)=56.20/58=0.969moln(O2)=m(O2)/M(O2)=85.30/32=2.666moln(C4H10)/1 and n(O2)/6.5 0.969/1 and 2.666/6.5 0.969 and 0.43The mass of water is calculated by the deficiency, by n(O2)n(O2)/n(H2O)=6.5/52.666*5=6.5*n(H2O)n(H2O)=2.05molV(H2O)=n(H2O)*Vm=2.05*22.4=45.92l=45920mlρ(H2O)=1g/mlm(H2O)=ρ*V(H2O)=1*45920=45920gAnswer - 45920g
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Pentane (C5H12) burns with oxygen to produce carbon dioxide and water. What mass of
C5H12 is required to produce 450.0 g of CO2? (3 marks)
C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l)
M(C5H12) = 72.15 g/mol;M(CO2) = 44.01 g/mol;n=m/M;n(CO2) = 450 g/(44.01 g/mol) = 10.22 moles;Proportion according to the reaction:1 mole (pentane) — 5 moles (CO2)x moles (pentane) — 10.22 moles (CO2)x...
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If 4.97 mol of C5H12 reacts with excess O2, how many moles of CO2 will be produced by the following combustion reaction?
C5H12+8O2⟶6H2O+5CO2
From the equation it is seen that from one mol of C5H12 five moles of CO2 is formed. So, the number of moles of CO2 may be found from a...
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Determine the amount of theoretical oxygen and theoretical air needed to burn
a.) 60 kg of butane (C4H10)
b.) 20 kg of propane (C5H12)
c.) 80 kg of octane (C8H18)
A. 2C4H10 + 13O2 ----> 8CO2 + 10H2OMolar mass of C4H10 = (12x4)+10= 58gmol-1molar mass of O2= 16x2=32gmol-12(58)g of C4H10 required 13(32)g of OxygenTherefore 60Kg of C4H10 will require 416x60/116...
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