Answer:
CaCO3 is the limiting reactant
55 g of CO2 is made
Explanation:
First we must put down the reaction equation;
CaCO3(s) + 2HCl(aq) ---------> CaCl2(s) + H2O(l) + CO2(g)
Number of mole of CaCO3 = 125g/100gmol-1 = 1.25 moles
From the reaction equation;
1 mole of CaCO3 yields 1 mole of CO2
Hence 1.25 moles of CaCO3 yields 1.25 moles of CO2
For HCl;
number of moles of HCl = 125g/36.5 g mol-1 = 3.42 moles
From the reaction equation;
2 moles of HCl yields 1 mole of CO2
3.42 moles of HCl yields 3.42 * 1/2 = 1.71 moles of CO2
Hence CaCO3 is the limiting reactant.
Mass of CO2 produced = 1.25g * 44 gmol-1 = 55 g of CO2
The expected speed is v = 85.5 km/h
v = 85.5 km/h = (85.5 km/h)*(0.2778 (m/s)/(km/h)) = 23.75 m/s
If there is an uncertainty of 2 meters in measuring the position, then within a 1-second time interval:
The lower measurement for the speed is v₁ = 21.75 m/s,
The upper measurement for the speed is v₂ = 25.75 m/s.
The range of variation is
Δv = v₂ - v₁ = 4 m/s
The uncertainty in measuring the speed is
Δv/v = 4/23.75 = 0.1684 = 16.84%
Answer: 16.8%
The balanced chemical reaction is written as :
Na2CO3<span> + 2HCl === 2NaCl + H2O + CO2
</span>
We are given the amount of NaCl to be produced from the reaction. This will be the starting point for the calculations. We do as follows:
120 g NaCl ( 1 mol / 58.44 g) ( 1 mol Na2CO3 / 2 mol NaCl)( 105.99 g / 1 mol ) = 1108.82 g Na2CO3 needed
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