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3 years ago

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For the reaction between ammonium phosphate and lead (IV) nitrate, producing ammonium nitrate and lead (IV) phosphate, how many

moles of lead (IV) nitrate would be needed to produce 11.75 moles of ammonium nitrate

1 answer:

NemiM [27]3 years ago

3 0

Answer:

2.938 mol Pb(NO₃)₄

Explanation:

Step 1: Given data

Moles of NH₄NO₃ to be produced: 11.75 mol

Moles of Pb(NO₃)₄: ?

Step 2: Write the balanced equation

4 (NH₄)₃PO₄ + 3 Pb(NO₃)₄ = Pb₃(PO₄)₄ + 12 NH₄NO₃

Step 3: Calculate the number of moles of Pb(NO₃)₄required to produce 11.75 moles of NH₄NO₃

The molar ratio of Pb(NO₃)₄ to NH₄NO₃ is 3:12.

11.75 mol NH₄NO₃ × (3 mol Pb(NO₃)₄/12 mol NH₄NO₃) = 2.938 mol Pb(NO₃)₄

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In the following compound (HCl) How many electrons are gained and lost by each atom?

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In HCL, one positive atom is given to chlorine from hydrogen so that it can complete it's octate. chlorine take one electron from hydrogen.

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‏‏‎ ‎

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A 1.24g sample of a hydrocarbon, when completely burned in an excess of O2 yields 4.04g Co2 and 1.24g H20. Draw plausible struct

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Answer:

Plausible structure has been given below

Explanation:

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Answer: $2.12\times 10^{25}$ atoms of hydrogen are there in

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There are $2.12\times 10^{25}$ atoms of hydrogen are there in

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3 years ago