Can someone explain how to do these chemistry problems?
Mg(s) + HCl ---> MgCl2(aq) + H2(g)
2) how many grams of chlorine gas must be reacted with excess sodium iodide if 10.0g of sodium chloride are needed?
NaI(aq) + Cl2(g) ---> NaCl(aq) + I2(s)
3) how many grams of oxygen are produced in the decomposition of 5.00 g of potassium chlorate?
KClO3(s) ----> KCl(s) + O2(g)
how do i do these types of problems? thank you very much
Answer:
For these type of problems, you need to follow the same basic steps.
1. Write a balanced equation
2. Determine if you have a limiting reactant, if it isn't already apparent. Use that to determine yields of products.
3. Do the conversion from mass---> moles ---> moles ---> mass by using the molecular weights and the stoichiometric ratios.
So, given that:
1) Mg + 2 HCl ---> MgCl2 + H2
20.0 g Mg * 1 mole/24.3 g * 1 mole MgCl2/mole Mg * 95.3 g/mole = 78.4 g MgCl2
2) 2 NaI + Cl2 ---> 2 NaCl + I2
10.0 g NaCl * 1 mole/58.5 g * 1 mole Cl2/2 mole NaCl * 71.0 g/mole = 6.07 g Cl2
3) 2 KClO3 ---> 2 KCl + 3 O2
5.00 g KClO3 * 1 mole/122.5 g * 3 mole O2/2 mole KClO3 * 32.0 g/mole = 1.96 g O2
stoichiometry
1) change given grams to moles using the mass from the periodic table
2) use the balanced equation to get the mole ratio of the chemicals
3) change back to grams
1) 20 g Mg / 23.4 g per mol = .823 mol Mg
2) 1 mol Mg : 1 mol MgCl2, so you produce 1 x .823 mol MgCl2
3) 1 x .823 = .823 mol MgCl2 so: .823 mol MgCl2 x 95.3 g/mol = 78.4 g MgCl2 produced
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