1) molar masses of O2 and NO2
O2: 32 g/mol
NO2: 14 + 32 = 46 g/mol
4 mol NO2 = 4mol * 46g/mol = 184 g
1 mol O2 = 32 g
184 g NO2 / 32 g O2 = 23 g NO2 / x g O2 => x = 23 g NO2 * [ 32 g O2 / 184 g NO2] = 4 g O2.
and 28.75 is 28.75/46+0.625 moles
Which equates to 0.625 x 32= 20 g oxygen. (1mole of O2 is 2x16=32 grams)
But oxygen/NO2 ratio is 1:4 so 1/4x20=5g produced
The reaction is
2Pb(NO₃)₂(s) → 2PbO(s) + 4NO₂(g) + O₂(g)
Moles (mol) = mass (g) / molar mass (g/mol)
Mass of NO₂ = 11.5 g
Molar mass of NO₂ = 46.00 g/mol
Moles of NO₂ = 11.5 g / 46.00 g/mol
= 0.25 mol
The stoichiometric ratio between NO₂ and O₂ is 2 : 1.
moles of O₂ formed = moles of formed NO₂ / 2
= 0.25 / 2
= 0.125 mol
Molar mass of O₂ = 32 g/mol
Hence, mass of O₂ = moles x molar mass
= 0.125 mol x 32 g/mol
= 4.0 g
As can be seen from the balanced chemical equation, 2 moles of lead nitrate produce 4 moles of nitrogen dioxide.
of lead nitrate produces of nitrogen dioxide.
184 g of nitrogen dioxide will be produced by 662.4 g of lead nitrate
So 11.5 g of nitrogen dioxide will be produced by= of lead nitrate
As can be seen from the balanced chemical equation, 2 moles of lead nitrate produce 1 mole of oxygen.
of lead nitrate produces 32 g of oxygen.
41.4 g of lead nitrate produces = of oxygen.
0.8 g of O₂
The balance chemical equation is as follow,
2 Pb(NO₃)₂ → 2 PbO + 4 NO₂ + O₂
According to equation,
184 g (4 moles) NO₂ is produced along with = 32 g (1 mole) O₂ gas
4.60 g NO₂ when formed will produce along = X g of O₂
Solving for X,
X = (32 g × 4.60 g) ÷ 184 g
X = 0.8 g of O₂
Using the balanced combustion equation you have written, it can be seen that for every mole of methane burned, one mole of carbon dioxide is produced. Thus, if 150 moles of carbon dioxide are produced, 150 moles of methane are being used.
Now, we need to convert that to grams. Using the molecular weight of methane as 16.0425 g/mol:
(16.0425 g/mol) * (150 mol) = 2,406.375 grams of methane
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Mole ratio: 4NO2 to 1O2
Number of moles of O2=0.250/4 =0.0625mol
Mass of O2 produced= 0.0625x32 =2g
find the moles of NO2 = mass/molar mass
= 4.60g/46 g/mol = 0.1 moles
2Pb(NO3)2 = 2PbO +4 NO2 +O2
by use of mole ratio between No2 to O2 which is 4:1 the moles of O2
= 0.1 x1/4 = 0.025moles
mass of O2 = moles x molar mass
= 0.025 moles x 32 g/mol = 0.8 grams of oxygen
11.5 g NO2 (1 mol NO2 / 46.0055 g NO2) = 0.250 mol NO2
We use the mole ratio of O2 and NO2 from their coefficients in the balanced chemical equation
2Pb(NO3)2(s) → 2PbO(s) + 4NO2(g) + O2 (g)
which is one mole O2 is to react with four moles of NO2, to compute for the number of moles of oxygen:
0.250 mol NO2 (1 mol O2 / 4 mol NO2) = 0.0625 mol O2
We can now calculate for the mass of oxygen O2 using its molar mass:
0.0625 mol O2 (31.998 g O2 / 1 mol O2) = 2.0 grams O2
Therefore, 2.0 grams of oxygen is produced.
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the equation below shows the decomposition of lead nitrate. how man...