Problems 1-7 are based on the following equation: NayC03(s) + 2 HCI(aq) 2 NaCl(s

Question

Problems 1-7 are based on the following equation: NayC03(s) + 2 HCI(aq) 2 NaCl(s) + H20(g) + CO200 1. First, check to determine if the reaction is balanced! Once balanced, calculate the mass (in grams) of NaCI produced by the reaction of 7.53 g Na COs with concentrated HCl. 2. Determine the grams of water produced if 23.4 g of HCl is allowed to react with excess Na2CO3? 3. If the reaction produced 3.42 x 102 molecules of CO2 gas, how many grams of HCI was reacted? 4. If 65.5 kg of NaCl is produced, how many moles of H0 will be produced? 5. How many moles of HCl are needed to completely react with 43.4 g of NazCOs? Determine the number of moles of HCl that are needed to produce 4.8 moles of CO2. 6. 7. If 2.3 moles of Na COs are allowed to react completely, how many formula units of NaCI will be produced?

Explanation / Answer

Since the reaction is balanced, we can proceed further.

1. Moles of NA2CO3 = 7.53/(molar mass) = 7.53gm/106gm.

Now moles of nacl is 2 times the moles of na2co3 as the balanced coefficient is 2. So moles of nacl = 0.07151

Mass pf nacl = mole× molar mas ( 23+35.5) = 0.07151× 58.5 = 4.18gm.

2. Now mole of HCL = 23.4/36.5 = 0.6411 mol.

Therefore moles of H20 produced by 0.6411 mol OF HCL will be half of HCL that is 0.6411/2 = . Mass of H20 = 0.6411/2 × 18 = 5.7699gm.

3. Mole of CO2 produced = 3.42 × 1024 / 6.022 × 1023 = 5.679 mol. therefore mol of HCL will be twice of CO2 i.e = 11.35 mol . Mass of HCL = 11.35×36.5 = 414.58gm.

4. moles of nacl produced = 65.5 /58.5 = 1.11 kilomoles moles of H20 produced will be half of NaCl , = 0.559 kilomoles or 559 moles

5. Moles of Na2CO3 to be reacted = 43.4/106 = 0.409 mol. Moles of Hcl needed is twice of that = 0.8188 moles.

6. Moles of CO2 = 4.8 , therefore moles of Hcl = 2 × 4.8 = 9.6 moles

7. Moles of Na2co3 = 2.3 moles , therefore moles of nacl = 2 × 2.3 = 4.6 moles will be produced.

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