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  • 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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  1. anonymous
    • one year ago
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    m1= 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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