First, write the balanced reaction. H 50 + 2NaOH ? Na2SO4 +2H 0 2. If 30g of sul

Question

First, write the balanced reaction. H 50 + 2NaOH ? Na2SO4 +2H 0 2. If 30g of sulfuric acid and 30g of sodium hydroxide are reacted, what is the limiting reactant? How many grams of sodium sulfate is formed? Chemical calculations – calculating products formed a. Fill the blanks for the single replacement reaction, 2K + CuCl2 __ _ ? 2KCI + Cu g 1.5 mol Molar ratios for this reaction is 2 mol 1 mol ? 2 mol Imol 1.5 molCu* 2 mol K Cl2 ? 3 mol K * 39.1g K 1mol Cucl2 So in order to make 1.5 mol of Cu 1. moicu 1 mol Cut morcu 1 mol Cu 1 mol Cu - 1.5 molCu demo 1 mol Cu 1.5 molCu*2 mol Kci 1 mol Cu 3 mol K 1.5 mol CuCl2 ? 3 mol KCl, converting to grams 1.5 mol CuC12 * - 3 mol KCI *74.6 g k?i 1 mol K 1 mol Cuct2 T mol KCI 1 17.39 (201.8g 233.8g (these are the answers) Do the following. Fill in the blanks. 3. C + 2Cuð ? CO2 + 2Cu _ _g 1.0 mol

Explanation / Answer

2.)

H2SO4 + 2 NaOH ---> Na2SO4 + H2O

30 g of H2SO4 => 30 g / 98.08 g/mol => 0.3059 mol

30 g NaOH => 30 g/ 40 g/mol => 0.75 mol

1 mol of H2SO4 require 2 mol of NaOH

so 0.3059 mol H2SO4 required 2*0.3059 => 0.6118 mol of NaOH

but amount of NaOH presents is 0.75 mol.

Hence H2SO4 is limitting reagent.

moles of H2SO4 = moles of Na2SO4

so moles of Na2SO4 => 0.3059 mol

mass of Na2SO4 = 0.3059 mol x 142 g/mol => 43.44 g of Na2SO4.

3.) C + 2 CuO ---> CO2 + 2 Cu

Need to produce 1.0 mol of Cu.

1 mol of C and 2 mol of CuO produces 1 mol of CO2 and 2 mol of Cu.

So 0.5 mol of C and 1 mol of CuO can produce 0.5 mol of CO2 and 1 mol of Cu.

0.5 mol of C = 0.5 mol x 12 g/mol => 6.0 g

1 mol of CuO = 1 mol x 79.54 g/mol => 79.54 g

0.5 mol of CO2 = 0.5 mol x 44 g/mol => 22.0 g

C       +   2 CuO     ---> CO2 + 2 Cu

6.0 g         79.54 g           22.0 g      1 mol

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