Cao + H2O ---->Ca(OH)2
Calculate the number of each reactant and the moles of the product
that is
moles = mass/molar mass
The moles of CaO= 56.08g/ 56.08g/mol(molar mass of Cao)= 1mole
the moles of water= 36.04 g/18 g/mol= 2.002moles
The moles of Ca (OH)2=74.10g/74.093g/mol= 1mole
The mass of differences of reactant and product can be therefore
explained as
1 mole of Cao reacted completely with 1 mole H2O to produce 1 mole of Ca(OH)2. The mass of water was in excess while that of CaO was limited
The complete balanced chemical reaction is:
2 AgNO3 + Na2S --> 2 NaNO3 + Ag2S
First let us calculate the number of moles of AgNO3.
moles AgNO3 = 0.315 M * 0.035 L
moles AgNO3 = 0.011025 mol
From the reaction, 1 mole of Na2S is needed for every 2
moles of AgNO3 hence:
moles Na2S required = 0.011025 mol AgNO3 * (1 mol Na2S / 2
mol AgNO3)
moles Na2S required = 5.5125 x 10^-3 mol
Therefore volume required is:
volume Na2S = 5.5125 x 10^-3 mol / 0.260 M
<span>volume Na2S = 0.0212 L = 21.2 mL</span>
Answer:
b)4.46 L/hr
Explanation:
To solve this question we need to convert the mL to liters (Using the conversion of 1000mL = 1L) and convert the time from seconds to hours (3600s = 1hr)
<em>mL to L:</em>
1.24mL/s * (1L / 1000mL) = 0.00124L/s
<em>seconds to hours:</em>
0.00124L/s * (3600s / 1hr) = 4.46L/hr
Right answer is:
<h3>b)4.46 L/hr
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