Question

Please I need help ASAP

Lead nitrate decomposes on heating as indicated in Equation. 2Pb(NO3)2(s) 2PbO(s) + 4NO₂(g) + O₂(g) (4.8) If a volume of 112 cm³ of oxygen gas was collected at STP when a sample of lead nitrate was completely decomposed by heating, calculate the; (a) mass of the lead nitrate sample. (b) mass of lead(II) oxide produced. (c) Volume of nitrogen dioxide gas produced at STP. (Pb=207, N = 14, O=16; molar volume of gas at STP = 22.4 dm³)​

Answer 1

To solve this problem, we'll need to use stoichiometry and the molar ratios from the balanced chemical equation. Here's how you can calculate the values:

(a) Mass of the lead nitrate sample:

From the balanced equation, we can see that 2 moles of lead nitrate (Pb(NO3)2) produce 1 mole of oxygen gas (O2). We know that the volume of oxygen gas collected is 112 cm³, which is equal to 112/1000 = 0.112 dm³ (converting cm³ to dm³).

According to the molar volume of gas at STP (22.4 dm³), 1 mole of any gas occupies 22.4 dm³ at STP. Therefore, the number of moles of oxygen gas can be calculated as:

moles of O2 = volume of O2 / molar volume at STP

moles of O2 = 0.112 dm³ / 22.4 dm³/mol = 0.005 mol

Since 2 moles of lead nitrate produce 1 mole of oxygen gas, we can determine the number of moles of lead nitrate as:

moles of Pb(NO3)2 = 2 * moles of O2

moles of Pb(NO3)2 = 2 * 0.005 mol = 0.01 mol

To calculate the mass of the lead nitrate sample, we'll use its molar mass:

mass of Pb(NO3)2 = moles of Pb(NO3)2 * molar mass of Pb(NO3)2

mass of Pb(NO3)2 = 0.01 mol * (207 g/mol + 2 * 14 g/mol + 6 * 16 g/mol)

mass of Pb(NO3)2 = 0.01 mol * 331 g/mol

mass of Pb(NO3)2 = 3.31 g

Therefore, the mass of the lead nitrate sample is 3.31 grams.

(b) Mass of lead(II) oxide produced:

From the balanced equation, we can see that 2 moles of lead nitrate (Pb(NO3)2) produce 2 moles of lead(II) oxide (PbO). So, the number of moles of PbO produced is equal to the number of moles of Pb(NO3)2.

mass of PbO = moles of PbO * molar mass of PbO

mass of PbO = 0.01 mol * (207 g/mol + 16 g/mol)

mass of PbO = 0.01 mol * 223 g/mol

mass of PbO = 2.23 g

Therefore, the mass of lead(II) oxide produced is 2.23 grams.

(c) Volume of nitrogen dioxide gas produced at STP:

From the balanced equation, we can see that 2 moles of lead nitrate (Pb(NO3)2) produce 4 moles of nitrogen dioxide gas (NO2). So, the number of moles of NO2 produced is twice the number of moles of Pb(NO3)2.

moles of NO2 = 2 * moles of Pb(NO3)2

moles of NO2 = 2 * 0.01 mol = 0.02 mol

Using the molar volume of gas at STP, we can calculate the volume of nitrogen dioxide gas:

volume of NO2 = moles of NO2 * molar volume at STP

volume of NO2 = 0.02 mol * 22.4 dm³/mol = 0.448 dm³

Therefore, the volume of nitrogen dioxide gas