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anonymous
 one year ago
In the following reaction, how many grams of lead(II) nitrate Pb(NO3)2 will produce 425 grams of sodium nitrate (NaNO3)?
Pb(NO3)2(aq) + 2 NaBr(aq) PbBr2(s) + 2 NaNO3(aq)
The molar mass of Pb(NO3)2 is 331.21 grams and that of NaNO3 is 85 grams.
828.03 grams
54.54 grams
218.14 grams
3312.10 grams
110.00 grams
anonymous
 one year ago
In the following reaction, how many grams of lead(II) nitrate Pb(NO3)2 will produce 425 grams of sodium nitrate (NaNO3)? Pb(NO3)2(aq) + 2 NaBr(aq) PbBr2(s) + 2 NaNO3(aq) The molar mass of Pb(NO3)2 is 331.21 grams and that of NaNO3 is 85 grams. 828.03 grams 54.54 grams 218.14 grams 3312.10 grams 110.00 grams

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0m1= 425 grams of sodium nitrate (NaNO3) m2= ? grams of lead(II) nitrate Pb(NO3)2 MM1= 85 grams of NaNO3 is MM2=molar mass of Pb(NO3)2 is 331.21 grams 1) convert the grams of sodium nitrate to moles of sodium nitrate ( n1=m1/MM1) n=moles: m1=mass of sodium nitrate; MM1= molecular mass of sodium nitrate. 2) according to the stoiquiometry of the reaction you can see that 1 mol of Pb(NO3)2 will produce 2 mol of NaNO3(aq). Then divide n1 by 2 to calculate the n2 or moles of Pb(NO3)2 that you will need. n2= n1/2 3) convert the moles of Pb(NO3)2 to grams with the same formula used in 1) but now isolate m. => m2= n2/MM2
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